Question

A 15.7 kg block is dragged over a rough, horizontal surface by a constant force of 83.1 N
acting at an angle of 25.7
◦
above the horizontal. The block is displaced 62.6 m, and the
coefficient of kinetic friction is 0.161.

Find the work done by the 83.1 N force.
The acceleration of gravity is 9.8 m/s
2
.
Answer in units of J.
013 (part 2 of 2) 10.0 points
Find the magnitude of the work done by the
force of friction.
Answer in units of J.

Answer 1

Answer:

The work done by the 83.1 N force is 4687.5 J.

The magnitude of the work done by the  force of friction is 1187.5 J.

Explanation:

Given:

Mass of the block, $m=15.7$ kg

Force acting on it, $F=83.1$ N

Angle of application of force, $\theta = 25.7$°

Displacement of the block, $d=62.6$ m

Coefficient of friction, $\mu =0.161$

Acceleration due to gravity, $g=9.8$ m/s²

Work done by a force is given as:

$Work,W=F\times d\times \cos\theta$

So, work done by the constant force is given as:

$W_{force}=83.1\times 62.6\times \cos(25.7)\\W_{force}=4687.5\textrm{ J}$

Now, in order to find work done by friction, we need to evaluate friction.

For the vertical direction, the net force is zero as there is no vertical motion.

Therefore,

$N + F\sin\theta = mg\\ N = mg-F\sin\theta$

Frictional force is given as:

$f=\mu N=\mu (mg-F \sin \theta)=0.161\times ((15.7\times 9.8)-(83.1\times \sin(25.7))=18.97\textrm{ N}$

Now, friction acts in the direction opposite to the displacement. Thus, angle between frictional force and displacement is 180°.

Therefore, work done by friction is:

$W_{fric}=f\times d\times \cos\theta_f\\W_{fric}=12.72\times 62.6\times \cos(-180)\\W_{fric}=18.97\times 62.6\times -1\\W_{fric}=-1187.5\textrm{ J}$

Negative sign indicates that frictional force is acting opposite to motion.

So, the magnitude of the work done by the  force of friction is 1187.5 J.