Question

I need a step by step on 1 and 2 so I can teach myself.

1. A solid, uniform sphere with a mass of 2.0 kg is rolling from rest down an incline plane from the top of the plane. The incline plane makes an angle of 20° with the horizontal and has a height of 2.0 m. At the bottom of the incline plane, the surface levels out to a frictionless horizontal surface. A spring with a spring constant of k is located 5.0 m down the horizontal surface. If the spring is compressed by 5.0 cm upon impact with the sphere, what is the spring constant? 2. A box with a mass of 5.0 kg is sliding from rest down an incline plane from the top of the plane. The incline is frictionless plane makes an angle of 20° with the horizontal and has a height of 2.0 m. At the bottom of the incline plane, the surface levels out to a horizontal surface with a kinetic coefficient of friction of 0.2. A spring with a spring constant of k = 10 N/m is located 5.0 m down the horizontal surface. How much is the spring compressed? 3. A 22,000 kg truck is heading down a 3.50° slope at 20.0 m/s. At this point, the driver realizes he has a break failure. A runaway ramp happens to be located 600.0 m ahead. The ramp slopes upward at an angle of 10.0°, and the coefficient of rolling friction between the tires and the gravel is r = 0.40. Ignore air resistance, and friction on the pavement. (a) How far along the ramp does the truck travel before stopping? (b) By how much does the thermal energy of the truck and ramp increase as the truck stops? 4. A block of mass m slides down a plane of angle 8. Find the speed of the block after it descended through a height h, assuming that it starts from rest and that the coefficient of friction is constant. 5. A 7.80 g bullet moving at 530 m/s penetrates a tree trunk to a depth of 5.40 cm. (a) Use work and energy considerations to find the average frictional force that stops the bullet. (b) Assuming the frictional force is constant, determine how much time elapses between the moment the bullet enters the tree and the moment it stops moving. 6. An oversized yo yo is made from two identical solid disks each of mass M = 1.80 kg and radius R = 9.7 cm. The two disks are joined by a solid cylinder of radius T = 4.00 cm and mass m = 1.00 kg as in the figure below. Take the center of the cylinder as the axis of the system, with positive torques directed to the left along this axis. All torques and angular variables are to be calculated around this axis. Light string is wrapped around the cylinder, and the system is then allowed to drop from rest. How long does it take the system to drop 1.10 m from rest?

Answer 1

Hi,

Hope you are doing well.

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PART 1:

Here we have to use Law of conservation of energy.

The sphere is initially at a height of 2 m. Hence it has some gravitational potential energy.

This energy will be converted into kinetic energy of rolling and finally into the potential energy stored in the spring.

Given that,

Mass of the sphere, m = 2 kg.

Initial height, h = 2 m.

Compression of the spring, x = 5 cm = 0.05 m

According to Law of conservation of energy.

mgh = -K 12

Where K is spring constant.

Therefore, spring constant of the spring is given by,

2mgh K= 2 = 2 x 2 kg x 9.8 m/s2 x 2 m (0.05 m)

::K = 78.4 J 0.0025 m2

:K = 31360 N/m

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PART 2:

Here also we have to use Law of conservation of energy.

The box is initially at a height of 2 m. Hence it has some gravitational potential energy.

This energy will be converted initially into kinetic energy(while sliding down the incline) and finally into the potential energy stored in the spring and for the work done to overcome friction.

Given that,

Mass of the box, m = 5 kg.

Initial height, h = 2 m.

Spring constant of the spring,

K = 10 N/m

Length of horizontal surface, d= 5 m.

Frictional force,

F=uk N = Mimg = 0.2 x 5 kg x 9.8 m/s2

F = 9.8 N

According to Law of conservation of energy.

P'A + z2 Y = ybu

Where x is compression of the spring.

Therefore, compression of the spring is given by,

2 (mgh – F.d) – 9.8 N x 5 m) 2 x (5 kg x 9.8 m/s2 x 2 m 10 N/m

98 J 10 N/m

::1 = V9.8 m

::x= 3.13 m

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