1. The weight in the following diagram has a mass of 0.750 kg and the cart has a mass of 0.52 kg. There is a friction force of 2.1 N acting on the cart. What is the tension in the string?

Question

a. 4.4 N
b. 4.3 N
c. 4.1 N
d. 4.2 N

2. An 8.5 kg block is attached to a weight. The coefficient of static friction between the block and the table is 0.71. What is the minimum mass required for the weight in order to start the block in motion?
a. 6.0 kg
b. 7.0 kg
c. 4.0 kg
d. 5.0 kg

For #1 I got 4.04N, but that isn't an option. I used f=ma where f=2.1N and m=0.52kg

For #2 i think the answer is A.

Answer 1

There should be a pulley between the weight and the cart/block.
Is the weight free-hanging, or is it on an incline?

If it is free-hanging, you need to calculate the acceleration a.
Then equate the solve for the tension by equating the tension that will cause an acceleration of both the weight and the cart/block.

Answer 2

yes, there is a pulley between the weight and the block

Answer 3

I understand that there should be a pulley, but is the weight free-hanging?
Can you describe the diagram in words?

In the following calculation for #1, I assume the weight is free-hanging from the pulley.

Draw a free-body-diagram (FBD) for mass m1 (=0.52) on the table.
Write the equilibrium equation for the motion, assumed towards the pulley, and in direction of tension of string T.
We have
T-2.1-(m1)a=0 =>
T=(m1)a+2.1 ...........(1)

Now draw a FBD for the weight, which accelerates against tension T. The corresponding equilibrium equation is then
T+(m2)a-(m2)g=0 ..........(2)

Eliminate a from (1) and (2)
(T-2.1)/(m1)=((m2)g-T)/(m2)

Cross multiply and solve for T, which gives
T=4.2496

Note:
The value of T=4.2496 is from using g=9.8 m/s&sup2, which rounds to one decimal place as 4.2.
Had the value of g used been 9.81, then T=4.2526, which rounds to 1 decimal place as 4.3.
So the answer is sensitive to the value of g assumed.