Worksheet: Power as the Rate of Work

In this worksheet, we will practice calculating the power of a moving body given that its motion and the force acting on it are represented in vector notation.

Q1:

A particle is moving under the action of the force . Its position vector at time is given by the relation . Find the rate of the work done by the force at .

Q2:

A particle of unit mass is moving under the action of a force 𝐹 = 2 𝑖 + 4 𝑗 . Its displacement, as a function of time, is given by 𝑆 ( 𝑡 ) = 2 𝑡 𝑖 + 7 𝑡 + 2 𝑡 𝑗 2 . Find the value of 𝑑 𝑑 𝑡 𝐹 𝑆 when 𝑡 = 4 .

Q3:

A body of mass 3 kg moves under the action of a force 𝐹 N. At time 𝑡 seconds, the velocity of the body is given by 𝑣 = ( 4 2 𝑡 ) 𝑖 + ( 5 2 𝑡 ) 𝑗 / s i n c o s m s . Find, in terms of 𝑡 , the power of the force, 𝐹 .

  • A ( 8 2 𝑡 + 1 0 2 𝑡 ) c o s s i n W
  • B ( 2 4 2 𝑡 + 1 0 2 𝑡 ) c o s s i n W
  • C ( 1 2 2 𝑡 + 1 5 2 𝑡 ) c o s s i n W
  • D ( 2 4 2 𝑡 + 3 0 2 𝑡 ) c o s s i n W
  • E ( 6 2 𝑡 + 3 0 2 𝑡 ) 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 𝑡 is given by the relation 𝑟 ( 𝑡 ) = 7 𝑡 𝑖 + 4 𝑡 𝑗 3 2 . Given that 𝐹 is measured in newtons, 𝑟 in metres, and 𝑡 in seconds, write an expression for the power of force 𝐹 at time 𝑡 .

  • A 3 3 3 2 𝑡 + 4 8 9 6 𝑡 3 W
  • B 7 4 9 7 𝑡 + 5 4 4 𝑡 3 W
  • C 1 6 6 6 𝑡 + 2 4 4 8 𝑡 3 W
  • D 1 4 9 9 4 𝑡 + 1 0 8 8 𝑡 3 W

Q5:

A particle of mass 4 g is moving under the action of two forces F and F , where F = ( 6 i + 3 j ) d y n e s and F = ( 3 i + 4 j ) d y n e s . The position vector of the particle is given as a function of time by r ( 𝑡 ) = 𝑎 𝑡 8 i + 𝑏 𝑡 + 1 j c m , where 𝑎 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 𝑡 seconds, is given by 𝑟 ( 𝑡 ) = 2 𝑡 𝑖 + 3 𝑡 𝑗 2 c m , where 𝑖 and 𝑗 are two perpendicular unit vectors. Given that the force 𝐹 was working at a rate of 35 erg/s when 𝑡 = 2 , and at 43 erg/s when 𝑡 = 4 , determine 𝐹 .

  • A 9 𝑖 + 𝑗 dynes
  • B 9 𝑖 + 𝑗 dynes
  • C 𝑖 9 𝑗 dynes
  • D 𝑖 + 9 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 𝑡 seconds is given by r i j = 9 𝑡 + 8 𝑡 . m Find the work done by the force F over the interval 0 𝑡 1 .

Q8:

A particle is moving under the action of the force . Its position vector at time is given by the relation . Find the rate of the work done by the force at .

Q9:

A particle of mass 2 g is moving under the action of two forces F and F , where F = ( 2 i + 3 j ) d y n e s and F = ( 4 i 7 j ) d y n e s . The position vector of the particle is given as a function of time by r ( 𝑡 ) = 𝑎 𝑡 1 i + 𝑏 𝑡 2 j c m , where 𝑎 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 𝑡 seconds, the velocity of the body is given by 𝑣 = ( 2 + 4 3 𝑡 ) 𝑖 + ( 2 + 4 3 𝑡 ) 𝑗 / s i n c o s m s . Find, in terms of 𝑡 , the power of the force, 𝐹 .

  • A ( 2 4 3 𝑡 + 2 4 3 𝑡 ) c o s s i n W
  • B ( 4 8 3 𝑡 + 2 4 3 𝑡 ) c o s s i n W
  • C ( 1 6 3 𝑡 + 1 6 3 𝑡 ) c o s s i n W
  • D ( 4 8 3 𝑡 + 4 8 3 𝑡 ) c o s s i n W
  • E ( 2 4 3 𝑡 + 4 8 3 𝑡 ) 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 𝑡 is given by the relation 𝑟 ( 𝑡 ) = 8 𝑡 𝑖 + 5 𝑡 𝑗 3 2 . Given that 𝐹 is measured in newtons, 𝑟 in metres, and 𝑡 in seconds, write an expression for the power of force 𝐹 at time 𝑡 .

  • A 1 0 2 4 𝑡 + 1 8 0 0 𝑡 3 W
  • B 2 3 0 4 𝑡 + 2 0 0 𝑡 3 W
  • C 5 1 2 𝑡 + 9 0 0 𝑡 3 W
  • D 4 6 0 8 𝑡 + 4 0 0 𝑡 3 W

Q12:

A particle of unit mass is moving under the action of a force 𝐹 = 𝑖 + 5 𝑗 . Its displacement, as a function of time, is given by 𝑆 ( 𝑡 ) = 7 𝑡 𝑖 + 5 𝑡 𝑡 𝑗 2 . Find the value of 𝑑 𝑑 𝑡 𝐹 𝑆 when 𝑡 = 4 .

Q13:

A constant force 𝐹 , measured in dynes, is acting on a body. The displacement of the body, after 𝑡 seconds, is given by 𝑟 ( 𝑡 ) = 𝑡 𝑖 5 𝑡 𝑗 2 c m , where 𝑖 and 𝑗 are two perpendicular unit vectors. Given that the force 𝐹 was working at a rate of 71 erg/s when 𝑡 = 7 , and at 31 erg/s when 𝑡 = 2 , determine 𝐹 .

  • A 3 𝑖 + 4 𝑗 dynes
  • B 3 𝑖 + 4 𝑗 dynes
  • C 4 𝑖 + 3 𝑗 dynes
  • D 4 𝑖 3 𝑗 dynes

Q14:

A constant force 𝐹 , measured in dynes, is acting on a body. The displacement of the body, after 𝑡 seconds, is given by 𝑟 ( 𝑡 ) = 2 𝑡 𝑖 6 𝑡 𝑗 2 c m , where 𝑖 and 𝑗 are two perpendicular unit vectors. Given that the force 𝐹 was working at a rate of 6 erg/s when 𝑡 = 6 , and at 14 erg/s when 𝑡 = 7 , determine 𝐹 .

  • A 7 𝑖 + 2 𝑗 dynes
  • B 7 𝑖 + 2 𝑗 dynes
  • C 2 𝑖 7 𝑗 dynes
  • D 2 𝑖 + 7 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 𝑡 seconds is given by r i j = 6 𝑡 + 5 𝑡 . m Find the work done by the force F over the interval 0 𝑡 1 .

Q16:

A body of mass 8 kg is moving under the action of the force F measured in newtons. Its position vector after 𝑡 seconds is given by r i j = 2 𝑡 + 6 𝑡 . m Find the work done by the force F over the interval 0 𝑡 1 .

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