# A horizontal force of 156N is used to push a 36.0 kg packing crate a distance of 7.00 m on a rough horizontal surface. /p>The acceleration of gravity is 9.81 m/s^2. If the crate moves with...

A horizontal force of 156N is used to push a 36.0 kg packing crate a distance of 7.00 m on a rough horizontal surface.

/p>

The acceleration of gravity is 9.81 m/s^2. If the crate moves with constant velocity, calculate

a.) the work done by the force. Answer in units of J.

b.) the coefficient of kinetic friction.

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### 2 Answers

Let M be the coefficient of friction between the surface and the surface on which it moves.

The forces acting on the crate are 36Kg weight force of it balanced by an equal and opposite normal reaction. The frictional force = M times normal reaction = M*mg balanced by the 156 N horizontal push.

Therefore, the net force on the crate is -M*(36 kg*g)+156 N =0 as there is no acceleration. Therefore, M= 156N/(36*9*81 N)=0.4417

The work done by the force of 156N= force*distance = 156N*7m=1092 Nm= 1092J

Given:

Mass of packing crate = m = 36 kg

Force to move crate at constant speed = f = 156 N

Distance moved = s = 7 m

Acceleration due to gravity = g = 9.81 m/s^2

We have to find out (1) work done by the force and (2) coefficient of kinetic friction.

Calculating work done:

Work done = Force*Distance = f*s = 156*7 = 1092 J

Calculating coefficient of kinetic friction:

Coefficient of kinetic friction is force required to keep the the crate in constant motion divided bu normal force. That is

Coefficient of kinetic friction = f/fn

Where fn is the normal force acting at right angle to the direction of motion and of frictional force. This is the force acting on the crate due to acceleration due to gravity. Thus

fn = m*g = 36*9.81 = 353.16

Therefore: Coefficient of kinetic friction = f/fn = 156/353.16 = 0.4417

Answer:

Work done by the force is 1092 J

Coefficient of kinetic friction is 0.4417