Question

1. Calculate the centripetal force for a 0.50 kg ball on a string whirled in a horizontal circle. The string is 0.50 m long. The ball makes 1 revolution per second.
2. A spring with a spring constant of 60 N/m is compressed by a 20 N force.       a) How much was the spring compressed? b) How much work was done on the spring?
3. A wooden crate sits on a wooden floor (µ_s = 0.6). The crate’s mass is 120 kg and a worker wants to move it along the floor. What’s the minimum force it will take to start the crate moving?
4. Find the resultant (magnitude and direction) of these forces: 300 N at 0⁰ and 400 N at 150⁰.
5. After falling from a cliff that is 100 m high, a 30 kg rock hits the ground below. Neglecting air resistance, how fast was it moving just as it hit?
6. How much heat will it take to completely melt 2 kg of lead that is at 227⁰ C?
7. What is the mutual attractive gravitational force between two people, each with a mass of 60 kg standing 2 m apart?
8. A wheel rotates at 24 revolutions every 3 minutes. This is equivalent to A) 16π rad/s    B) 50.2 rad/s    C) 0.837 rad/s          D) 1.19 rad/s

Answer 1

1. Solution :

Mass of the ball, m = 0.50 kg

Length of the string = radius of the horizontal circle = 0.50 m

radius of the horizontal circle, r = 0.50 m

Number of revolutions per second, n = 1

Angular speed of the ball,

Linear speed of the ball,

Centripetal force of the ball,

Therefore, the centripetal force of the ball, .

2. Solution:

Spring constant of the spring,

Compression force on the spring,

(a) Spring Force is given by, , where is the elongation or compression in the spring from its natural length.

Compression produced in the spring,

So, the spring was compressed by 0.33 m.

(b) Work done on the spring,

So, the work done on the spring, W = 3.27 J

3. Solution :

Coefficient of static friction,

Mass of the crate, m = 120 kg

The static friction acts in the direction opposite to the direction of motion of the crate. The worker pushes the crate with a force in the direction opposite to the direction of static frictional force.  

As there is no vertical motion, normal force(N) due to floor will balance the weight(mg) of the crate, i.e.

Minimum force required to move the crate is equal to the static frictional force(Fs) which opposes the motion of the crate.

Therefore, the minimum force which will set the crate in motion is 705.6 N.