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.

Answer 1

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

$20.89 = 13a$

$a = 1.61 m/s^2$

Answer 2

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