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A force F at an angle θ above the horizontal is used to pull a heavy suitcase of weight mg a distance d along a level floor at constant velocity. The coefficient of friction between the floor and the suitcase is µ. The work done by the frictional force is:
(A) -Fd cos θ
(B) -µ Fd cos θ
(C) -µmgd
(D) -µ mgd cos θ

Respuesta :

Answer:

option A

Explanation:

given,

Force = F

angle = θ

weight on suitcase = mg

distance = d

constant velocity so, acceleration a = 0

coefficient of friction = µ

Work done = ?

Work done is equal to force into displacement.

Friction act opposite to the force acting so, work done by frictional force will be negative.

frictional force will act into horizontal direction opposite to force.

here displacement is equal to d

now,

W = -F d cos θ

Hence,the correct answer is option A

snehashish65

The work done by the frictional force between the flor and suitcase is [tex]-Fdcos \theta[/tex].  Hence, option (A) is correct.

Given data:

Magnitude of force is, F.

Angle of inclination is, [tex]\theta[/tex].

Weight of suitcase is, [tex]W = mg[/tex].

Coefficient of friction between the floor and suitcase is, [tex]\mu[/tex].

The work done by frictional force is given as,

[tex]W =-f_{f} \times d[/tex] ......................................(1)

Negative sign denotes that friction tends to resist the motion of suitcase. Which means it is acting in the direction opposite to the motion of suitcase.

The frictional force is caused by the horizontal component of force [tex](Fcos \theta)[/tex] to produce constant velocity of suitcase.

Then equation (1) is expressed as,

[tex]W =-Fcos \theta \times d\\W =-Fdcos \theta[/tex]

Thus, we can conclude that  work done by the frictional force is [tex]-Fdcos \theta[/tex]. And option (A) is correct.

Learn more about frictional force here:

https://brainly.com/question/4230804?referrer=searchResults

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