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In this lesson, we will learn how to calculate 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 β πΉ = β 2 β π + 4 β π . Its displacement, as a function of time, is given by β π ( π‘ ) = 2 π‘ β π + οΉ 7 π‘ + 2 π‘ ο β π 2 . Find the value of π π π‘ οΊ β πΉ β β π ο when π‘ = 4 .

Q3:

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

Q4:

A body of mass 17 kg moves under the action of a force β πΉ . Its position vector at time π‘ is given by the relation β π ( π‘ ) = οΉ 7 π‘ ο β π + οΉ 4 π‘ ο β π 3 2 . Given that πΉ is measured in newtons, π in metres, and π‘ in seconds, write an expression for the power of force β πΉ at time π‘ .

Q5:

A particle of mass 4 g is moving under the action of two forces β πΉ 1 and β πΉ 2 , where β πΉ = οΊ 6 β π + 3 β π ο 1 d y n e s and β πΉ = οΊ 3 β π + 4 β π ο 2 d y n e s . The position vector of the particle is given as a function of time by β π ( π‘ ) = ο οΉ π π‘ β 8 ο β π + οΉ π π‘ + 1 ο β π ο 2 2 c m , where π and π 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 β πΉ , measured in dynes, is acting on a body. The displacement of the body, after π‘ seconds, is given by β π ( π‘ ) = οΊ 2 π‘ β π + 3 π‘ β π ο 2 c m , where β π and β π are two perpendicular unit vectors. Given that the force β πΉ was working at a rate of 35 erg/s when π‘ = 2 , and at 43 erg/s when π‘ = 4 , determine β πΉ .

Q7:

A body of mass 5 kg is moving under the action of the force F measured in newtons. Its position vector after π‘ seconds is given by r i j = οΉ 9 π‘ + 8 π‘ ο . ο© ο¨ m Find the work done by the force F over the interval 0 β€ π‘ β€ 1 .

Q8:

Q9:

A particle of mass 2 g is moving under the action of two forces β πΉ 1 and β πΉ 2 , where β πΉ = οΊ β 2 β π + 3 β π ο 1 d y n e s and β πΉ = οΊ β 4 β π β 7 β π ο 2 d y n e s . The position vector of the particle is given as a function of time by β π ( π‘ ) = ο οΉ π π‘ β 1 ο β π + οΉ π π‘ β 2 ο β π ο 2 2 c m , where π and π 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 β πΉ N. At time π‘ seconds, the velocity of the body is given by β π£ = ο ( 2 + 4 3 π‘ ) β π + ( β 2 + 4 3 π‘ ) β π ο / s i n c o s m s . Find, in terms of π‘ , the power of the force, β πΉ .

Q11:

A body of mass 4 kg moves under the action of a force β πΉ . Its position vector at time π‘ is given by the relation β π ( π‘ ) = οΉ 8 π‘ ο β π + οΉ 5 π‘ ο β π 3 2 . Given that πΉ is measured in newtons, π in metres, and π‘ in seconds, write an expression for the power of force β πΉ at time π‘ .

Q12:

A particle of unit mass is moving under the action of a force β πΉ = β π + 5 β π . Its displacement, as a function of time, is given by β π ( π‘ ) = β 7 π‘ β π + οΉ 5 π‘ β π‘ ο β π 2 . Find the value of π π π‘ οΊ β πΉ β β π ο when π‘ = 4 .

Q13:

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

Q14:

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

Q15:

A body of mass 3 kg is moving under the action of the force F measured in newtons. Its position vector after π‘ seconds is given by r i j = οΉ 6 π‘ + 5 π‘ ο . ο© ο¨ m Find the work done by the force F over the interval 0 β€ π‘ β€ 1 .

Q16:

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

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