Question

The box moves at a constant speed. If the mass of the box is 6.4 kg, it is pushed 3.6 m vertically upward, the coefficient of friction is 0.35, and the angle θ is 30.0°, determine the following. (a) the work done (in J) on the box by F (b) the work done (in J) on the box by the force of gravity J (c) the work done (in J) on the box by the normal force J (d) the increase in gravitational potential energy (in J) of the system of the box and earth as the box moves up the wall

Answer 1

Part A.

Work-done is given by:

W = Applied Parallel Force*displacement

W = Fp*d

Fp = Applied parallel Force

Since given that box is moving at constant speed, which means net acceleration of the box is zero, So from Newton's 2nd law Net Force on the box in vertical direction will also be zero. Now in vertical direction there are only two forces, one is applied force's parallel component and other one is gravity, So for net force to be zero, both of these will be equal and in opposite direction. Which means

Fp = m*g = 6.4*9.81

Wp = 6.4*9.81*3.6

Wp = 226.0 J

Part B.

Work-done by gravity

Wg = Fg*d

d = displacement = 3.6 m

Fg = -m*g = -6.4*9.81 (Negative sign because force and displacement are in opposite direction)

So,

Wg = -6.4*9.81*3.6 = -226.0 J

Part C.

Work-done by Normal force

W = N*d

Since Normal force will be perpendicular to displacement, So Work-done by Normal Force will be zero.

(If we take parallel component of normal force than it will be zero)

Part D.

Increase in GPE will be given by:

GPE = -Wg

GPE = -(-226.0)

GPE = 226.0 J

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