Anyone who can explain how to do these questions, your help is much appretiated!

1) If you use a horizontal force of 30.0N to slide a 12.0kg wooden crate across a floor at a constant velocity, what is the coefficient of sliding friction between crate and floor?

The answer is supposed to be 0.255


2) The maximum force a grocery sack can withstand and not rip is 250N. If 20kg of groceries are lifted from the floor to the table with an acceleration of 5m/s squared, will the sack hold?

The answer is supposed to be that the sack will rip.

3) A force of 40N accelerates a 5.0kg block at 6.0m/s squared along a horizontal suface.
a) How large is the frictional force?
b) What is the coefficiant of friction?

The answers are supposed to be a) 10N
b) 0.20
Thank you again:D

1) Because you are pulling at a constant force, and have caused the object to move at a CONSTANT velocity, the sum of the forces acting on the 12.0kg crate is zero.

If there is a horizontal force 30N pushing the crate, then there must be a horizontal force of 30N against the crate (an object can still be moving even if the forces are 0, it just wouldn't accelerate!)

So the frictional force must be equal to 30N in the opposite direction.

Frictional force for a moving object is often denoted as

F(friction) = (coeff.fric)*F(normal), We can also see here that the coefficent is unitless!

F(normal) is just the normal force that the crate exerts to the ground, which is equal and opposite to the force due to gravity (mg).

F(normal) = mg = 12.0 *9.8 = 117.6N

F(friction) = mu*F(normal), mu is the greek letter that represents friction

mu = F(friction)/F(normal) = 30N/117.6N =0.255

2) The forces in play here are gravity, and the force used to accelerate the bag upwards!

If the maximum force the bag can hand is 250N, we need to find the force of 20kg of groceries exerts on the bag.

Because the bag is accelerated upwards, against gravity, then the force exerted to life the bag must be greater than gravity!

Gravity accelerates objects at a rate of -9.8m/s^2, so an acceleration of 5m/s^2 would be a net change of 14.8 m/s^2.

F=m*a, we know mass m, and we know the overall acceleration a to be 14.8

F=20*14.8 which is indeed greater than 250, and therefore the bag would rip

3 I will be a little more vague in answering this because I do want you to understand the concepts in these problems!

for question 3(a and b), utilize newton's third and second law, it is very similar to question 1.

For question 3, b I got 0.33333 when the answer is supposed to be 0.20
Anyone know how to get the right answer?

3) A force of 40N accelerates a 5.0kg block at 6.0m/s squared along a horizontal suface.
a) How large is the frictional force?
b) What is the coefficiant of friction?

We have to look at all the forces in play

First we have frictional force
Then there is the force accelerating the block. Because the block is accelerating, the force used is greater than frictional force.

Remember from above that frictional force is equal to mu * normal force or
Therefore the force applied must be greater than frictional force

F(friction)-F(applied) = -ma , It is minus because the forces run in opposite directions, it really depends on YOUR coordinate system
F(friction) + -40 = 5*0*-6.0
F(friction) = 10N

Now for part B
F(friction) = mu* F(normal)
10 = mu*m*g
10 = mu*5*9.8 (If you had used an acceleration of 6.0 here, you would end up with 0.333.. however, normal force deals only with acceleration in the Y direction)
mu = 0.204

I hope this helps

Hm... I'm sorry, I have to disagree with the first sentence you said:


1) Because you are pulling at a constant force, and have caused the object to move at a CONSTANT velocity...

Constant force leads to constant acceleration, from Newton
s equation F=ma. The force is constant, yes, but the force due to friction is equal to that force (like you said later), meaning that the resultant force is zero. When force is zero, no acceleration occurs anymore, but constant velocity is the result.

It is that way that you have to say that.

Good eye, I didn't catch that, my mistake. I meant to point out that a constant velocity meant the forces had to be equal and opposite, thanks!