Question

A 3 kg block is being pulled across a table by a horizontal force of 80 N. The block is also experiencing a frictional force of 5 N. Draw a force diagram.

a) [Diagram]
b) [No Diagram]
c) [Diagram Not Provided]
d) [Force Diagram Not Applicable]

What is the acceleration of the block?
a) 25 m/s²
b) 15 m/s²
c) 10 m/s²
d) 5 m/s²

If the block started from rest and was pushed for 5 seconds, what is its final velocity?
a) 50 m/s
b) 25 m/s
c) 20 m/s
d) 15 m/s

Answer 1

The 3 kg block has a net force of 75 N (80 N pull minus 5 N friction), which results in an acceleration of 25 m/s². After being pushed for 5 seconds, its final velocity would be 125 m/s, which doesn't match any of the given answers, indicating a possible typo or error.

Step-by-step explanation:

To calculate the acceleration of the block, we need to consider the net force acting on it. Since there is an applied force of 80 N and a frictional force of 5 N opposing it, the net force is 75 N.

Using Newton's second law, F = ma, where F is the net force, m is the mass, and a is the acceleration:

a = F/m = 75 N / 3 kg = 25 m/s²

The final velocity v after being pushed for t seconds can be found using the equation v = at.

So, v = 25 m/s² * 5 s = 125 m/s. However, none of the options provided match this calculation. The student may need to double-check the choices or the question itself as there might be a typo.

Answer 2

1. A force diagram of a 3 kg block is being pulled across a table by a horizontal force of 80 N and is also experiencing a frictional force of 5 N is Option A.

2. The acceleration of the block is 25 m/s² (Option A).

3. The final velocity of the block is 125 m/s (None correct option).

Step-by-step explanation:

The force diagram for the block being pulled across the table would show a horizontal force of 80 N in the direction of the pull, a gravitational force of 30 N downward, and a frictional force of 5 N in the opposite direction of the pull.

To find the acceleration of the block, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the net force acting on the block is the difference between the applied force and the frictional force.

Net force = Applied force - Frictional force

Net force = 80 N - 5 N = 75 N

Using the formula F = ma, where F is the net force, m is the mass of the block (3 kg), and a is the acceleration, we can rearrange the formula to solve for the acceleration:

a = F / m

a = 75 N / 3 kg = 25 m/s²

The acceleration of the block is 25 m/s².

To find the final velocity of the block after being pushed for 5 seconds from rest, we can use the equation of motion:

v = u + at

where v is the final velocity, u is the initial velocity (0 m/s), a is the acceleration (25 m/s²), and t is the time (5 seconds).

Substituting the values into the equation, we get:

v = 0 + 25 m/s² * 5 s = 125 m/s

The final velocity of the block is 125 m/s.

Thus, the correct option is

1. A (The diagram for the question nuber 1 can be seen in the attachment).

2. A.

3. None correct option.