Worksheet: Power as the Rate of Work

In this worksheet, we will practice defining the power of a force as the derivative of the work done by the force.

Q1:

A particle is moving under the action of the force Fij=8+5. Its position vector at time 𝑑 is given by the relation rij(𝑑)=ο€Ή3π‘‘βˆ’π‘‘βˆ’4+ο€Ό52𝑑+10𝑑+5. Find the rate of the work done by the force at 𝑑=3.

Q2:

A particle of unit mass is moving under the action of a force Fij=βˆ’2+4. Its displacement, as a function of time, is given by Sij(𝑑)=2𝑑+ο€Ή7𝑑+2π‘‘ο…οŠ¨. Find the value of dd𝑑(β‹…)FS when 𝑑=4.

  • A60 power units
  • B228 power units
  • C476 power units
  • D124 power units

Q3:

A body of mass 3 kg moves under the action of a force F N. At time 𝑑 seconds, the velocity of the body is given by vij=[(βˆ’4βˆ’2𝑑)+(5βˆ’2𝑑)]/sincosms. Find, in terms of 𝑑, the power of the force, F.

  • A ( βˆ’ 6 2 𝑑 + 3 0 2 𝑑 ) c o s s i n W
  • B ( 2 4 2 𝑑 + 1 0 2 𝑑 ) c o s s i n W
  • C ( 8 2 𝑑 + 1 0 2 𝑑 ) c o s s i n W
  • D ( 1 2 2 𝑑 + 1 5 2 𝑑 ) c o s s i n W
  • E ( 2 4 2 𝑑 + 3 0 2 𝑑 ) c o s s i n W

Q4:

A body of mass 17 kg moves under the action of a force F. Its position vector at time 𝑑 is given by the relation rij(𝑑)=ο€Ή7𝑑+ο€Ή4π‘‘ο…οŠ©οŠ¨. Given that 𝐹 is measured in newtons, π‘Ÿ in meters, and 𝑑 in seconds, write an expression for the power of force F at time 𝑑.

  • A ο€Ή 3 , 3 3 2 𝑑 + 4 , 8 9 6 𝑑   W
  • B ο€Ή 7 , 4 9 7 𝑑 + 5 4 4 𝑑   W
  • C ο€Ή 1 4 , 9 9 4 𝑑 + 1 , 0 8 8 𝑑   W
  • D ο€Ή 1 , 6 6 6 𝑑 + 2 , 4 4 8 𝑑   W

Q5:

A particle of mass 4 g is moving under the action of two forces F and F, where Fij=(6+3)dynes and Fij=(3+4)dynes. The position vector of the particle is given as a function of time by rij(𝑑)=ο‘ο€Ήπ‘Žπ‘‘βˆ’8+𝑏𝑑+1ο…οοŠ¨οŠ¨cm, where π‘Ž and 𝑏 are constants. Determine, in ergs per second, the power at which the resultant force acts on the particle 8 seconds after the start of motion.

Q6:

A constant force F, measured in dynes, is acting on a body. The displacement of the body, after 𝑑 seconds, is given by rij(𝑑)=ο€Ή2𝑑+3π‘‘ο…οŠ¨cm, where i and j are two perpendicular unit vectors. Given that the force F was working at a rate of 35 erg/s when 𝑑=2, and at 43 erg/s when 𝑑=4, determine F.

  • A ( 9 + ) i j dynes
  • B ( βˆ’ 9 + ) i j dynes
  • C ( βˆ’ 9 ) i j dynes
  • D ( + 9 ) i j dynes

Q7:

A body of mass 5 kg is moving under the action of the force F measured in newtons. Its position vector after 𝑑 seconds is given by rij=ο€Ή9𝑑+8𝑑.m Find the work done by the force F over the interval 0≀𝑑≀1.

Q8:

A particle is moving under the action of the force Fij=4+. Its position vector at time 𝑑 is given by the relation rij(𝑑)=ο€Ό72π‘‘βˆ’3𝑑+3+ο€Ή8π‘‘βˆ’3𝑑+4ο…οŠ¨οŠ¨. Find the rate of the work done by the force at 𝑑=7.

Q9:

A particle of mass 2 g is moving under the action of two forces F and F, where Fij=(βˆ’2+3)dynes and Fij=(βˆ’4βˆ’7)dynes. The position vector of the particle is given as a function of time by rij(𝑑)=ο‘ο€Ήπ‘Žπ‘‘βˆ’1+ο€Ήπ‘π‘‘βˆ’2ο…οοŠ¨οŠ¨cm, where π‘Ž and 𝑏 are constants. Determine, in ergs per second, the power at which the resultant force acts on the particle 6 seconds after the start of motion.

Q10:

A body of mass 2 kg moves under the action of a force F N. At time 𝑑 seconds, the velocity of the body is given by vij=[(2+43𝑑)+(βˆ’2+43𝑑)]/sincosms. Find, in terms of 𝑑, the power of the force, F.

  • A ( 2 4 3 𝑑 + 4 8 3 𝑑 ) c o s s i n W
  • B ( 4 8 3 𝑑 + 2 4 3 𝑑 ) c o s s i n W
  • C ( 2 4 3 𝑑 + 2 4 3 𝑑 ) c o s s i n W
  • D ( 1 6 3 𝑑 + 1 6 3 𝑑 ) c o s s i n W
  • E ( 4 8 3 𝑑 + 4 8 3 𝑑 ) c o s s i n W

Q11:

A body of mass 4 kg moves under the action of a force F. Its position vector at time 𝑑 is given by the relation rij(𝑑)=ο€Ή8𝑑+ο€Ή5π‘‘ο…οŠ©οŠ¨. Given that 𝐹 is measured in newtons, π‘Ÿ in meters, and 𝑑 in seconds, write an expression for the power of force F at time 𝑑.

  • A ο€Ή 1 , 0 2 4 𝑑 + 1 , 8 0 0 𝑑   W
  • B ο€Ή 2 , 3 0 4 𝑑 + 2 0 0 𝑑   W
  • C ο€Ή 4 , 6 0 8 𝑑 + 4 0 0 𝑑   W
  • D ο€Ή 5 1 2 𝑑 + 9 0 0 𝑑   W

Q12:

A particle of unit mass is moving under the action of a force Fij=+5. Its displacement, as a function of time, is given by Sij(𝑑)=βˆ’7𝑑+ο€Ή5π‘‘βˆ’π‘‘ο…οŠ¨. Find the value of dd𝑑(β‹…)FS when 𝑑=4.

  • A38 power units
  • B188 power units
  • C373 power units
  • D4 power units

Q13:

A constant force F, measured in dynes, is acting on a body. The displacement of the body, after 𝑑 seconds, is given by rij(𝑑)=ο€Ήπ‘‘βˆ’5π‘‘ο…οŠ¨cm, where i and j are two perpendicular unit vectors. Given that the force F was working at a rate of 71 erg/s when 𝑑=7, and at 31 erg/s when 𝑑=2, determine F.

  • A ( βˆ’ 3 + 4 ) i j dynes
  • B ( 3 + 4 ) i j dynes
  • C ( 4 + 3 ) i j dynes
  • D ( 4 βˆ’ 3 ) i j dynes

Q14:

A constant force F, measured in dynes, is acting on a body. The displacement of the body, after 𝑑 seconds, is given by rij(𝑑)=ο€Ή2π‘‘βˆ’6π‘‘ο…οŠ¨cm, where i and j are two perpendicular unit vectors. Given that the force F was working at a rate of 6 erg/s when 𝑑=6, and at 14 erg/s when 𝑑=7, determine F.

  • A ( 7 + 2 ) i j dynes
  • B ( βˆ’ 7 + 2 ) i j dynes
  • C ( 2 βˆ’ 7 ) i j dynes
  • D ( 2 + 7 ) i j dynes

Q15:

A body of mass 3 kg is moving under the action of the force F measured in newtons. Its position vector after 𝑑 seconds is given by rij=ο€Ή6𝑑+5𝑑.m Find the work done by the force F over the interval 0≀𝑑≀1.

Q16:

A body of mass 8 kg is moving under the action of the force F measured in newtons. Its position vector after 𝑑 seconds is given by rij=ο€Ή2𝑑+6𝑑.m Find the work done by the force F over the interval 0≀𝑑≀1.

Q17:

The power of a machine is given by the relation 𝑃=3𝑑+7, where 𝑑 is the time elapsed in seconds. Find the work done by the machine in the first 8 seconds.

Q18:

The power of an engine is given by ο€Ό8π‘‘βˆ’115π‘‘οˆοŠ¨ hp, where π‘‘βˆˆ[0,103] is the time in seconds. Find the power of the engine π‘ƒοŠ§ when 𝑑=36seconds, the work done π‘Š over the time interval [0,12], and the maximum power 𝑃max of the engine.

  • A 𝑃 = 2 0 1 . 6  h p , π‘Š = 5 3 7 . 6 β‹… k g - w t m , 𝑃 = 1 , 2 0 0 m a x h p
  • B 𝑃 = 2 0 1 . 6  h p , π‘Š = 4 0 , 3 2 0 β‹… k g - w t m , 𝑃 = 2 4 0 m a x h p
  • C 𝑃 = 4 8 9 . 6  h p , π‘Š = 5 3 7 . 6 β‹… k g - w t m , 𝑃 = 7 2 0 m a x h p
  • D 𝑃 = 5 4 7 . 2  h p , π‘Š = 7 7 , 7 6 0 β‹… k g - w t m , 𝑃 = 8 4 0 m a x h p

Q19:

The power of an engine at time 𝑑 seconds is given by 𝑃(𝑑)=ο€Ή12𝑑+8π‘‘ο…οŠ¨W. Find the work done by the engine between 𝑑=2s and 𝑑=3s.

Q20:

A car has mass 1,066 kg. At time 𝑑 seconds, its engine works at a rate of 𝑃=𝑑+9π‘‘ο…οŠ¨W. Given that at 𝑑=6s the car’s speed is 78 km/h , find its speed at 𝑑=10s. Give your answer to the nearest m/s.

Q21:

Find the time taken for a car of 1,236 kg to reach a speed of 126 km/h, given that the car started from rest and that the power of the engine is constant and equal to 103 metric horsepower. Use 𝑔=9.8/ms.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.