A man pushing a crate of mass m = 92.0 kg at a speed of v = 0.870 m/s encounters

Question

A man pushing a crate of mass m = 92.0 kg at a speed of v = 0.870 m/s encounters a rough horizontal surface of length = 0.65 m as in the figure below. If the coefficient of kinetic friction between the crate and rough surface is 0.350 and he exerts a of 278 N on the crate constant horizontal force (a) Find the magnitude and direction of the net force on the crate while it is on the rough surface. magnitude direction …Sel ---Select- (b) Find the net work done on the crate while it is on the rough surface. (c) Find the speed of the crate when it reaches the end of the rough surface. m/s

Explanation / Answer

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given mass of the crate = m = 92 kg

speed of the crate = v = 0.870 m/s

length l = 0.65 m

Coeficient of friction = 0.35

Horizontal force Fh = 278 N

part A :

frictional force acting on the crate = Ff = mg    this acts opposite to the applied force

Ff = 0.35 * 92* 9.8 = 315.56 N

Applied force F = 278 N

Net Force = Fapplied - Ff

Net Force = 278 - 315.56   = -37.56 N , in a diretion opposite to the applied force.
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part B:

Net Work Done On Crate while it is on the Rough Surface is

W = Fnet * Distance

W = -37.56 * 0.65 = -24.14 Joules

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part C:

acceleration of the crate = a = FNet/m

a = -37.56/92 = -0.408 m/s^2

use the kinematic equation v^2 - u^2 = 2 a S

V^2 = (0.87^2) -(2 * 0.408 * 0.65)

V = 0.476 m/s is the Speed of the crate when it reaches the end of the rough surface

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