Question

A 16 kg block is dragged over a rough, horizontal surface by a constant force of 188 N acting at an angle of 33.2 ◦ above the horizontal. The block is displaced 96.1 m, and the coefficient of kinetic friction is 0.147. Find the magnitude of the work done by the force of friction.

Answer 1

Answer:

Magnitude of work done W = 2.214 kW.

Explanation:

Given :

Mass of object , m = 16 kg.

Angle from horizon , $\theta=33.2^o$.

Force applied , F = 188 N.

Displacement of block , D = 96.1 m.

Coefficient of kinetic friction , $\mu=0.147$ .

Component of force, required to move the block, $F\ cos\theta=188\times cos\ 33.2^o=157.31\ N.$

Also , frictional force , $f=\mu N=\mu (mg)=0.147\times 16\times 9.8=23.045\ N.$

We know, work done , $W = Fdcos\theta$   ( Here $\theta = 180^o$ , because the direction of displacement is opposite the direction of frictional force .)

Putting all value in above equation :

We get , $W=23.045\times 96.1\times cos\ 180^o=-2214.62 \ W=-2.214\ kW.$

Magnitude of work done W = 2.214 kW.

Hence, this is the required solution.

Answer 2

Answer:$W=-762.84\ J$

Explanation:

Given

mass of block $m=16\ kg$

Force $F=188\ N$

Force is applied at an angle of $\theta =33.2^{\circ}$

Displacement of block $s=96.1\ m$

coefficient of kinetic friction $\mu _k=0.147$

Friction force acting on block $f_r$

$f_r=\mu _kN$

where N=Normal reaction

$N=mg-F\sin \theta$

$N=16\times 9.8-188\times \sin (33.2)$

$N=156.8-102.94$

$N=53.859\approx 54\ N$

$f_r=0.147\times 54$

$f_r=7.938\ N$

Work done by friction

$W=f_r\cdot s$

$W=7.938\times 96.1\times \cos (180)$

$W=-762.84\ J$

Negative sign indicates that displacement is against the direction of force.