A man pushes his lawn mower across the horizontal ground. He pushes down and to

Question

A man pushes his lawn mower across the horizontal ground. He pushes down and to the right on the mower at an angle of 50 degrees from the horizontal. The mass of the mower is 40 kg. The coefficient of static friction of the mower with the ground is 0.2 and the coefficient of kinetic friction is 0.1. a) Draw a sketch of the situation and a force diagram of the mower, labeling all forces clearly. b) With what force does the man have to push to start the mower moving from rest? c) If he continues to push with this force after the mower is moving, what will be its acceleration? d) How long will he have to push the with this force unitil the it is moving at 1 m/s? e) Discuss whether or not your answers to parts b through d pass the "common sense" test -do not discuss your method. Be specific about why you think the answers do or do not make sense.

Explanation / Answer

Given that

Mass (m) =50kg

Degree(theta) =50degrees

us =0.2

uk =0.1

Here the horizonal force must be equal to the static force

FN= mg+Fsintheta

Fx=fs

Fcostheta =usFN

Fcostheta =us(mg+Fsintheta)

Fcos50 =(0.2)[40*9.81+Fsin50]

0.64F =78.4+0.153F

F =160N

c)

Here we will use kinetic friction force

FN=Fx-fk

ma =Fcostheta-ukFN

40kg*a =160cos50 -(0.10)[40*9.81+160sin50]

a =1.28m/s2

d)

We know that the kinetic equation of motion is

vf =vi+at

1m/s =0+(1.28)t

t =0.78sec

e)

It is correct because as it starts moving it wil gain 1m/s in 0.78sec

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