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?

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.

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

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.

1. Sorry, I'm not here to solve physics problems. But I can still make you smile! Why did the cart go to the comedy club? Because it wanted to lighten its load and have a good laugh!

2. Ah, physics again! Well, if you got A as the answer, go ahead and give yourself a pat on the back. But remember, laughter is also a great way to start things in motion! What do you call a funny weight lifting? Comic-rows!

For Question 1:

To find the tension in the string, we can use Newton's second law of motion, which states that force equals mass times acceleration (F = ma). In this case, the cart and the weight are connected by the string, so they will have the same acceleration.

Let's break down the forces acting on the system:
- Weight force (downwards) = mass * gravitational acceleration = 0.750 kg * 9.8 m/s^2 = 7.35 N
- Friction force (opposite to the direction of motion) = 2.1 N
- Tension force (in the direction of motion) = unknown

Considering the forces in the vertical direction, we can write:
Sum of forces vertically = Weight force - Tension force = 0.

Since the weight force and the tension force are equal (as there is no vertical acceleration), we know that the tension force is 7.35 N.

Now, considering the forces in the horizontal direction, we can write:
Sum of forces horizontally = Tension force - Friction force = mass * acceleration.

Plugging in the values, we have:
7.35 N - 2.1 N = 0.52 kg * acceleration.

Simplifying the equation:
Acceleration = (7.35 N - 2.1 N) / 0.52 kg = 5.25 N / 0.52 kg ≈ 10.0962 m/s^2.

Now, using the acceleration we found, we can find the tension force by applying Newton's second law:
Tension force = mass * acceleration = 0.750 kg * 10.0962 m/s^2 = 7.572 N.

So, the tension in the string is approximately 7.572 N, which is not listed as an option. Therefore, none of the provided options (a, b, c, d) are correct for Question 1.

For Question 2:

To find the minimum mass required for the weight to start the block in motion, we can use the concept of static friction. When the block is not moving, the static friction force will be equal to the applied force trying to move the block.

The formula for static friction force is given by:
Force of static friction = coefficient of static friction * normal force.

The normal force is the force exerted by the table on the block in the vertical direction, which is equal to the weight of the block:
Normal force = mass * gravitational acceleration = 8.5 kg * 9.8 m/s^2 = 83.3 N.

Plugging in the coefficient of static friction (0.71) and the normal force (83.3 N) into the formula, we have:
Force of static friction = 0.71 * 83.3 N ≈ 59.143 N.

To overcome the static friction force and start the block in motion, the applied force (the weight in this case) needs to exceed the force of static friction. Therefore, the minimum mass required for the weight can be found by dividing the force of static friction by the gravitational acceleration:
Minimum mass = Force of static friction / gravitational acceleration = 59.143 N / 9.8 m/s^2 ≈ 6.04 kg.

The closest option to the calculated minimum mass is 6.0 kg, so the answer for Question 2 is option A.

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.