homework. help me(Physics)

Question

An applied force of 250N is acting parallel to the level ground on a 40kg crate. There is a coefficient of friction of 0.38 between the surfaces during the motion. The crate starts from rest and reaches a velocity of 10m/s(forwards).

 

1)How much is the acceleration for the object? 2) How much work is done for this motion?

Answer 1

Draw the free body diagram:

Here, W is the weight of the crate,N is the normal force on the crate,F_k is the frictional force, and a is the acceleration of the object.

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Calculate the normal force by the equilibrium in vertical direction.

ΣF_y = 0:

N-W=0

N=W  =mg  =(40)(9.81)  =392.4 N

Calculate the frictional force:

F_k  = µ_k N  =(0.38)(392.4)=149.112 N

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Calculate the acceleration by using the quilibrium in horizontal direction.

ΣF_x = ma:

F-F_k=ma

250-149.112 = 40a

40a =100.888

a =2.522 m/s²

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Initial velocity, u =0

Final velocity ,v=10 m/s

v² -u² =2as

s = (v² -u²)/2a

  =(10² -0 ²)/2(2.522)

  =19.82 m

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Calculate the work done in the motion

W =(F-F_k  )s

    =(250-149.112)(19.82)   

    = 2000.16 J

    =2 kJ