Question

Jane works at a grocery store. She has a large crate of Bananas with a mass of 13 kg to move to the produce aisle.

a. What is the weight of the crate?

b. What is the normal force acting on the crate when it sits on the floor? Explain how you found the answer.

c. If she pushes on the box with a force of 40 N and there is no friciton, what is the acceleraiton of the box?

d. If the coefficient of static friction between the crate and the floor is 0.2, what si the maximum force of static friction acting on the crate?

e. If she pushes on the crate with a force of 12 N, what is the force of static friction on the crate ?Explain how you found the answer.

f. If she pushes on the create with a force of 40 N, and the coefficient of kinetic friction is 0.15, what is the net force on the create in the x-direction?

g. Based on your answer in part (f), what is the acceleration of the crate?

Answer 1

a) W=mg=127.4 N

b) When the crate is resting of the floor, the normal force is equal to the weight, so N=127.4 N

c) a=F/m=40/13=3.08 m/s/s

d) f=0.2N=25.48 N

e) From the answer above we see that the crate won't move unless the force prevails static friction. So in the current situation it's 12 N

f) 40-0.15N=20.89 N

g) a=F/m=20.89/13=1.61 m/s/s

Answer 2

PART a)

As we know that weight is product of mass and gravity

so here we have

`W = mg`

`W = 13(9.8) = 127.4 N`

Part B)

Normal force is counter balanced by weight of the crate

so here we have

N = W = 127.4 N

PART C)

As we know that

F = ma

now we have

m = 13 kg

F = 40 N

now we have

`40 = 13(a)`

`a = 3.08 m/s^2`

Part D)

Maximum static friction force is given as

`F_s = \mu_s N`

here we know that

`\mu_s = 0.2`

now we have

`F_s = 0.2(127.4) = 25.48 N`

PART E)

Since applied force on the block is

F = 12 N

Now since applied force is less than maximum static friction force

So it will not slide

So here friction force will be same as applied force

Static friction = 12 N

PART f)

Kinetic friction force is given as

`F_k = \mu_k N`

here we know that

`\mu_k = 0.15`

now friction force is

`F_k = 0.15(127.4) = 19.11 N`

now we have

`F_(net) = F - F_k`

`F_(net) = 40 - 19.11`

`F_(net) = 20.89 N`

Part g)

as we know that

F = ma

`20.89 = 13a`

`a = 1.61 m/s^2`