Worksheet: Work Done by a Force Expressed in Vector Notation

In this worksheet, we will practice calculating the work done by a constant force vector acting on a body over a displacement vector using the dot product.

Q1:

A particle moves in a plane in which i and j are perpendicular unit vectors. A force, Fij=(9+)N acts on the particle. The particle moves from the origin to the point with position vector (āˆ’9+6)ij m. Find the work done by the force.

Q2:

A force Fij=(š‘šāˆ’9)N acts on a particle, causing a displacement sij=[āˆ’5+(š‘š+6)]cm. If the work done by the force is 0.02 J, what is the value of š‘š?

Q3:

A particle moved from point š“(āˆ’7,āˆ’1) to point šµ(āˆ’4,6) along a straight line under the action of the force Fij=š‘Ž+š‘. During this stage of motion, the work done by the force was 106 units of work. The particle then moved from šµ to another point š¶(āˆ’8,āˆ’3) under the effect of the same force. During this stage of motion, the work done by the force was āˆ’138 units of work. Determine the two constants š‘Ž and š‘.

  • A š‘Ž = 1 2 , š‘ = 1 0
  • B š‘Ž = 1 2 , š‘ = āˆ’ 1 0
  • C š‘Ž = āˆ’ 1 2 , š‘ = 1 0
  • D š‘Ž = āˆ’ 1 2 , š‘ = āˆ’ 1 0

Q4:

A particle moved from point š“(āˆ’2,āˆ’2) to point šµ(6,10) along a straight line under the action of the force Fij=š‘˜āˆ’6 acting in the opposite direction to the displacement ļƒ š“šµ. Find the work done by the force F.

Q5:

A particle moved on a plane from the point š“(āˆ’8,6) to the point šµ(2,5) under the action of a force of 17 N whose line of action made an angle of šœƒ with the š‘„-axis where sinšœƒ=817. Find the work done by this force over the displacement ļƒ š“šµ.

  • A142 or āˆ’158 work units
  • B āˆ’ 6 5 or āˆ’95 work units
  • C āˆ’ 1 4 2 or 158 work units
  • D65 or 95 work units

Q6:

A particle moved from point š“(7,āˆ’3) to point šµ(āˆ’9,2) along a straight line under the action of a force F of magnitude 8āˆš10 N acting in the same direction as the vector cij=āˆ’3āˆ’. Calculate the work done by the force, given that the magnitude of the displacement is measured in meters.

Q7:

A body of mass 3 kg is moving under the action of a force F, such that its displacement sij(š‘”)=ļ€¹5š‘”ļ…+(7š‘”)ļŠØ. Find the work done by this force in the first 6 seconds of its motion, given that the displacement is measured in meters, the force in newtons, and the time š‘” in seconds.

Q8:

The displacement of a particle of mass 30 g is given as a function of time by the relation si(š‘”)=ļ€¹5š‘”āˆ’6š‘”ļ…ļŠØ, where i is a constant unit vector, š‘  is measured in centimeters, and š‘” in seconds. Given that the particle started its motion at š‘”=0, find the force š¹ acting on the particle and the work done š‘Š by this force during the first 7 seconds of motion.

  • A š¹ = āˆ’ 2 1 0 i d y n , š‘Š = 5 4 , 3 9 0 e r g
  • B š¹ = āˆ’ 3 5 5 i d y n , š‘Š = 9 1 , 9 4 5 e r g
  • C š¹ = āˆ’ 3 0 i d y n , š‘Š = 7 , 7 7 0 e r g
  • D š¹ = āˆ’ 3 6 0 i d y n , š‘Š = 9 3 , 2 4 0 e r g

Q9:

A particle moves in a plane in which i and j are perpendicular unit vectors. Its displacement from the origin at time š‘” seconds is given by rij=ļ‘ļ€¹2š‘”+7ļ…+(š‘”+7)ļļŠØm and it is acted on by a force Fij=(6+3)N. How much work does the force do between š‘”=2s and š‘”=3s?

Q10:

The position vector of a particle of mass 3 kg moving under the action of a force is given as a function of time š‘” by the relation rij(š‘”)=(āˆ’4š‘”āˆ’10)+(3š‘”+5)ļŠØļŠØ, where i and j are two perpendicular unit vectors. Calculate the work done by the force between š‘”=3 to š‘”=4.

Q11:

A body of mass 2 kg is moving under the action of three forces, FļŠ§, FļŠØ, and FļŠ©, where FijļŠ§=š‘āˆ’3, FijļŠØ=āˆ’4+3, and FijļŠ©=āˆ’10+š‘Ž, and i and j are two perpendicular unit vectors, š‘Ž and š‘ are constants, and each force is measured in newtons. The displacement of the body is expressed by the relation sij(š‘”)=ļ€¹4š‘”ļ…+ļ€¹3š‘”āˆ’8š‘”ļ…ļŠØļŠØ, where the displacement is measured in meters, and the time š‘” is in seconds. Determine the work done by the resultant of the forces in the first 6 seconds of motion.

  • A 3,120 J
  • B 1,584 J
  • C 3,024 J
  • D āˆ’ 7 6 8 J

Q12:

A force Fij=(āˆ’4āˆ’9) N is acting on a particle whose position vector as a function of time is given by rij(š‘”)=ļ€¹(āˆ’9š‘”āˆ’8)+ļ€¹āˆ’3š‘”+2ļ…ļ…ļŠØ m. Calculate the work š‘Š done by the force F between š‘”=3 and š‘”=8seconds.

Q13:

An object moves 10 m in the direction of ji+. There are two forces acting on this object: FijkļŠ§=+2+2 N and FijkļŠØ=5+2āˆ’6 N. Find the total work done on the object by the two forces.

Hint: You can take the work done by the resultant of the two forces or you can add the work done by each force.

  • A 3 0 āˆš 2 Nā‹…m
  • B 1 0 āˆš 2 Nā‹…m
  • C 5 0 āˆš 2 Nā‹…m
  • D40 Nā‹…m
  • E 6 0 āˆš 2 Nā‹…m

Q14:

An object moves 10 m in the direction of j. There are two forces acting on this object: FijkļŠ§=++2 N and FijkļŠØ=āˆ’5+2āˆ’6 N. Find the total work done on the object by the two forces.

Hint: You can take the work done by the resultant of the two forces or you can add the work done by each force.

Q15:

An object moves 20 meters in the direction of kj+. There are two forces acting on this object: FijkļŠ§=++2 N and FijkļŠØ=+2āˆ’6 N. Find the total work done on the object by the two forces.

  • A āˆ’ 2 0 āˆš 2 Nā‹…m
  • B 2 0 āˆš 2 Nā‹…m
  • C 5 0 āˆš 2 Nā‹…m
  • D āˆ’ 1 0 āˆš 2 Nā‹…m
  • E 1 0 āˆš 2 Nā‹…m

Q16:

A body moves under the force Fij=6āˆ’9 from point š“(āˆ’2,8) to point šµ(1,āˆ’7). Determine the work done by the force, where the displacement is measured in meters and the force in newtons.

Q17:

A force of Fik=6+4newtons acts on a body and moves it from point š“(āˆ’9,āˆ’2,āˆ’3) to point šµ(8,āˆ’7,āˆ’1). Determine the work done by the force F, where the displacement is measured in metres.

Q18:

A body moves from point š“(1,2,5) to point šµ(5,5,āˆ’4) under a force F. Determine the work done if the force has a magnitude of āˆš57 newtons and direction cosines ļ“’5āˆš5757,4āˆš5757,4āˆš5757ļ““, taking the displacement as measured in meters.

Q19:

A particle is moving in a straight line under the action of the force Fij=āˆ’8āˆ’3 from point š“(8,7) to point šµ(8,āˆ’5). Find the work done š‘Š by the force F.

  • A š‘Š = 3 2 units of work
  • B š‘Š = 3 6 units of work
  • C š‘Š = 9 6 units of work
  • D š‘Š = 9 6 units of work
  • E š‘Š = 1 3 4 units of work

Q20:

A body is displaced by Sijk=āˆ’9+2+10 m with a force Fijk=āˆ’9+2āˆ’3 N. What is the work done?

Q21:

A force of 8 newtons moves a body from š“(2,2,6) to šµ(3,āˆ’7,5). Measuring using meters, what is the work done?

  • A 8 āˆš 8 3 J
  • B24 J
  • C664 J
  • D āˆ’ 7 2 J

Q22:

A body moves in a plane in which i and j are perpendicular unit vectors. Two forces, FijļŠ§=(9āˆ’2)N and FijļŠØ=(9āˆ’7)N act on the body. The particle moves from the point with position vector (āˆ’6+2)ij m to the point (2+3)ij m. Find the work done by the resultant of the forces.

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