A physics TA pushes a 20.0-kg equipment cart at a constant velocity a distance o

Question

A physics TA pushes a 20.0-kg equipment cart at a constant velocity a distance of 16.0 m down the hall. The force she exerts on the cart is directed at an angle 21.0° below the horizontal and force of friction opposing the motion of cart is 60.0 N. Determine the following.

(a) magnitude of the force exerted on the cart by the TA

____________ N

(b) work done by the force the TA exerts on the cart

__________ J

(c) work done by the force of friction acting on the cart

_________ J

(d) work done by the force of gravity acting on the cart

________ J

Explanation / Answer

A) Since the velocity is constant, net force is zero.

Horizontal force of TA= 60 N

Total force by TA = 60/cos 21 degree = 64.27 N

B) Work done by TA = 64.27* cos 21 degree *16

= 960 J

C) work done by friction = - 60*16 = -960 J

D) work done by gravity = 0 N

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