Worksheet: Conservative and Nonconservative Forces

In this worksheet, we will practice calculating the position, velocity, and kinetic energy of objects that experience variable forces.

Q1:

Consider a particle on which several forces act, one of which is known to be constant in time: Fij=(3.0+4.0)N. As a result, the particle moves along the π‘₯-axis from π‘₯=0.0m to π‘₯=5.0m in some time interval. What is the work done by F?

Q2:

Consider a particle on which several forces act, one of which is known to be constant in time: Fij=(3.0+4.0)N. As a result, the particle moves along a straight path from a Cartesian coordinate of (0 m, 0 m) to (5.0 m, 6.0 m). What is the work done by F?

Q3:

A force 𝐹(π‘₯)=ο€Ήβˆ’5.0π‘₯+7.0π‘₯ο…οŠ¨N acts on a particle. How much work does the force do on the particle as it moves from π‘₯=2.0m to π‘₯=5.0m?

Q4:

A particle of mass 2.0 kg moves under the influence of the force 𝐹(π‘₯)=ο€Ήβˆ’5π‘₯+7π‘₯ο…οŠ¨N. If its speed at π‘₯=βˆ’4.0m is 𝑣=20.0/ms, what is its speed at π‘₯=4.0m?

Q5:

A particle moves along a curved path defined by 𝑦(π‘₯)=10Γ—(1+(0.1π‘₯))sin, from π‘₯=0m to π‘₯=10πœ‹m. It is subject to force of variable magnitude 𝐹(π‘₯)=10Γ—(0.1π‘₯)cos, which acts in the 𝑦-direction only. How much work is done by the force? (Give your answer to 3 significant figures.) ο„Έ(π‘Žπ‘₯)π‘₯=π‘₯2+14π‘Ž(2π‘Žπ‘₯)+𝐢.cosdsin Remember to set your calculator to radians mode.

Q6:

A 4.0-kg particle moving along the π‘₯-axis is acted upon by the force whose functional form appears in the figure. The velocity of the particle at π‘₯=0 is 𝑣=6.0/ms.

Find the particle’s speed at π‘₯=2.0m.

Find the particle’s speed at π‘₯=4.0m.

Find the particle’s speed at π‘₯=10.0m.

Q7:

A particle moving in the π‘₯𝑦-plane is subject to a force given by 𝐹(π‘₯,𝑦)=ο€Ή36⋅(π‘₯+𝑦)(π‘₯+𝑦),Nmij where π‘₯ and 𝑦 are in meters. With this force applied to the particle, it moves in a straight line from a point with the position (2.1+3.2)ij m to a point with the position (6.3+9.6)ij m. How much work is done on the particle by the force?

Q8:

A force Fij=((5.0𝑦)+(5π‘₯))/Nm acts on an object as the object moves along a parabolic curve, from a point with the position (0.0+0.0)ij m to a point with the position (0.0+3.3)ij m. Find the work done by the force.

Q9:

A force corresponds to the potential energy function π‘ˆ(π‘₯)=ο€Όβˆ’2.5π‘₯+1.2π‘₯ J. Find the force exerted at the position π‘₯=2.5 m.

Q10:

The force on a particle with a mass of 1.5 kg varies with position according to 𝐹(π‘₯)=ο€Ήβˆ’1.8π‘₯ο…οŠ¨ N. At a position π‘₯=2.8 m, the body’s speed is 3.7 m/s in the positive π‘₯-direction.

Calculate the mechanical energy of the particle, taking the origin as the point of zero potential.

Calculate the particle’s velocity at the position π‘₯=0.45 m, taking the origin as the point of zero potential.

Calculate the mechanical energy of the particle, taking the position π‘₯=3.0 m as the point of zero potential.

Calculate the particle’s velocity at the position π‘₯=0.45 m, taking the position π‘₯=3.0 m as the point of zero potential.

Q11:

A particle is acted on by several forces, one of which is known to be constant in time, Fij=(5.3+7.1) N. The particle moves first along the π‘₯-axis from π‘₯=0.0 m to π‘₯=3.5 m, and then parallel to the 𝑦-axis from 𝑦=0.0 m to 𝑦=2.2 m. What is the magnitude of the work done by F?

Q12:

How much work is done by a force given by 𝐹(π‘₯)=ο€Όβˆ’1.2π‘₯ N on a particle as the particle moves along the π‘₯-axis from π‘₯=1.3 m to π‘₯=3.6 m?

Q13:

Consider a particle on which a force acts that depends on the position of the particle, Fij=((3.2𝑦)+(2.7π‘₯)) N. Find the magnitude of the work done by this force when the particle moves along a parabolic path from a point with the position (0.0+0.0)ij to a point with the position (3.0+9.0)ij.

Q14:

A force 𝐹(π‘₯)=(βˆ’2.6π‘₯) N acts on a body of mass 0.62 kg. At a position π‘₯=2.8 m, the body s speed is 3.7 m/s. What is the speed of the body at the position π‘₯=1.2 m?

Q15:

A force 𝐹(π‘₯)=ο€Ό2.3π‘₯N acts on a particle as it moves along the positive π‘₯-axis.

How much work does the force do on the particle to move the particle from π‘₯=1.5m to π‘₯=4.6m?

Taking the potential energy of the particle to be zero at π‘₯=1.0m, find the potential energy of the particle at π‘₯=3.6m.

Q16:

You are in a room in a basement with a smooth concrete floor and a nice rug. The rug is 3 m wide and 4 m long. You have to push a very heavy box from one corner of the rug to its opposite corner. The magnitude of friction between the box and the rug is 55 N, but the magnitude of friction between the box and the concrete floor is only 40 N. Will you do more work against friction going around the floor or across the rug? How much extra work would it take?

  • AAround the floor, 280 J
  • BAcross the rug, 5 J
  • CAcross the rug, 275 J
  • DAround the floor, 5 J
  • EAround the floor, 275 J

Q17:

An 85.0 kg cross-country skier is climbing a 4.00∘ slope at a constant speed of 1.50 m/s and encounters air resistance of 20.0 N.

Find his power output for work done against the gravitational force and air resistance.

What average force does he exert backward on the snow to accomplish this?

If he continues to exert the same force backward on the snow and to experience the same air resistance when he reaches a level area, how long will it take him to reach a velocity of 10.0 m/s?

Q18:

The force Fij=(π‘Žπ‘₯𝑦+π‘Žπ‘¦π‘₯) N and the force Fij=𝑦π‘₯+2𝑦(π‘₯𝑏)ln N, where π‘Ž and 𝑏 are constants with appropriate units.

If 𝐴 is the derivative of the 𝑦-component of 𝐹 with respect to π‘₯ and 𝐡 is the derivative of the π‘₯-component of 𝐹 with respect to 𝑦, what is the ratio of 𝐴 to 𝐡?

  • A 𝑦 π‘₯
  • B π‘₯ 
  • C 𝑦 
  • D π‘₯ 𝑦
  • E1

If 𝐢 is the derivative of the 𝑦-component of 𝐹 with respect to π‘₯ and 𝐷 is the derivative of the π‘₯-component of 𝐹 with respect to 𝑦, what is the ratio of 𝐢 to 𝐷?

Q19:

A two-dimensional conservative force F has a magnitude of zero at all points on the π‘₯-axis and the 𝑦-axis and satisfies the condition 𝑦=ο€Ύπ‘₯=(4.0/)π‘₯𝑦ddddNmFFο—ο˜οŠ©. What is the magnitude of F at the point π‘₯=𝑦=1.0 m?

Q20:

The potential energy of a particle due to conservative forces acting on it is given by π‘ˆ(π‘₯)=14𝑐π‘₯οŠͺ, where 𝑐=8/Nm, and no nonconservative forces act on it. The particle is moving along the π‘₯-axis and its total energy at π‘₯=0 is 2 J. At a point 𝑃 on the positive π‘₯-axis, the particle has zero kinetic energy.

What is the value of 𝑃?

What magnitude force acts on the particle at 𝑃?

If the kinetic energy of the particle is 1 J at π‘₯=0m, what magnitude force acts on it at that point?

Q21:

A particle with a mass π‘š is suspended from a string of negligible mass and a length of 1.0 m, as shown in the diagram. The particle is displaced to a position where the taut string is at an angle of 30∘ from the vertical, and the particle is released from rest at that position. The particle moves through an arc, where the lowest point of the arc is the point 𝑃.

What is the instantaneous speed of the particle at point 𝑃?

What is the vertically upward displacement of the particle from point 𝑃 when its instantaneous speed is 0.81 m/s?

Q22:

A helicopter is hovering at an altitude of 1.00 km when a panel from its undercarriage comes loose and falls toward Earth, as shown in the diagram. Just before the panel hits the ground, it has a speed of 45.0 m/s. What energy 𝐸 was dissipated by atmospheric drag during the panel’s fall?

Q23:

The mechanical energy, 𝐸, of an object, is equal to the sum of its kinetic energy, 𝐾, and its potential energy, π‘ˆπΈ=𝐾+π‘ˆ.

The energy dissipated by non-conservative forces acting upon an object is the change in the mechanical energy, Δ𝐸=Ξ”(𝐾+π‘ˆ). A particle of mass 1 kg moves from point 𝐴 to point 𝐡, and is acted upon by a non-conservative force as it does so, losing energy. If the object had a mechanical energy of 4 J at point 𝐴, what value of potential energy, π‘ˆ, must it have had in order to move at a constant velocity of 2 m/s between points 𝐴 and 𝐡.

Q24:

The potential energy for a particle undergoing one-dimensional motion along the π‘₯-axis is π‘ˆ(π‘₯)=2(π‘₯βˆ’π‘₯)οŠͺ, where π‘ˆ is measured in joules and π‘₯ is measured in meters. The particle has a constant mechanical energy 𝐸=βˆ’0.25 J.

What positive region along the π‘₯-axis is the particle’s motion confined to?

  • A 0 . 2 4 ≀ π‘₯ ≀ 0 . 6 8 m m
  • B 0 . 1 5 ≀ π‘₯ ≀ 0 . 6 6 m m
  • C 0 . 4 3 ≀ π‘₯ ≀ 0 . 8 8 m m
  • D 0 . 5 4 ≀ π‘₯ ≀ 0 . 8 6 m m
  • E 0 . 3 8 ≀ π‘₯ ≀ 0 . 9 2 m m

What is the position of the stable equilibrium point on the positive π‘₯-axis for the particle?

What positive region along the π‘₯-axis is the particle’s motion confined to if its mechanical energy is a constant 0.25 J?

  • A 0 . 2 2 ≀ π‘₯ ≀ 1 . 0 m m
  • B 0 . 1 0 ≀ π‘₯ ≀ 1 . 2 m m
  • C 0 . 0 ≀ π‘₯ ≀ 1 . 1 m m
  • D 0 . 2 0 ≀ π‘₯ ≀ 1 . 2 m m
  • E 0 . 1 2 ≀ π‘₯ ≀ 1 . 1 m m

Q25:

A particle of mass 0.75 kg has a constant mechanical energy of 𝐸=5.0J and a potential energy given by π‘ˆ(π‘₯)=12π‘˜π‘₯. The particle is at rest at 𝑑=0.0s.

Find the position of the particle at 𝑑=2.5s if the value of π‘˜ is 15 N/m.

If the particle initially has zero potential energy, what is its initial speed?

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