1. As you and your bandmates prepare to rehearse, you push on your drum set to slide it into position. You exert a force of F = 24 N, kinetic friction opposes the motion with a force of Ff = 22 N, and you slide the drums a distance of d = 1.5 m. What is the net work done on the drums? What is the change in the box’s kinetic energy?
2. A 0.005 kg bullet hits a 0.5 kg block of stationary wood. If the bullet becomes lodged in the block and they fly off together with a velocity of 10.0 m/s, what was the original velocity of the bullet before the collision?
To find the answers to these questions, we will need to apply some principles from physics. Let's break down each question and go step by step.
1. To find the net work done on the drums, we need to calculate the work done by both the applied force and the frictional force.
a) Work done by the applied force: The work done by a force can be calculated using the formula W = F * d * cosθ, where W is the work done, F is the force applied, d is the distance the object moves, and θ is the angle between the force and the direction of motion.
Given:
Applied force, F = 24 N
Distance moved, d = 1.5 m
Since the applied force and the displacement are in the same direction, the angle θ between them is 0°. Therefore, cosθ = cos0° = 1.
Work done by the applied force = 24 N * 1.5 m * 1 = 36 J
b) Work done by the frictional force: The work done by the frictional force can be calculated using the same formula as above, but with the magnitude of the frictional force and the cosine of the angle between the frictional force and the direction of motion.
Given:
Frictional force, Ff = 22 N
Distance moved, d = 1.5 m
Since the frictional force opposes the motion, the angle between the force and the direction of motion is 180°. Therefore, cosθ = cos180° = -1.
Work done by the frictional force = 22 N * 1.5 m * (-1) = -33 J (Note the negative sign indicates work done against the motion)
Net work done on the drums = Work done by applied force + Work done by frictional force
Net work done = 36 J + (-33 J) = 3 J
The net work done on the drums is 3 Joules.
To find the change in the box's kinetic energy, we can use the work-energy theorem, which states that the net work done on an object is equal to its change in kinetic energy.
Change in kinetic energy = Net work done
Change in kinetic energy = 3 J
Therefore, the change in the box's kinetic energy is 3 Joules.
2. To find the original velocity of the bullet before the collision, we can use the principle of conservation of momentum.
According to the law of conservation of momentum, the total momentum before the collision should be equal to the total momentum after the collision. In this case, we have:
Initial momentum before the collision = Final momentum after the collision
The momentum of an object can be calculated as the product of its mass and velocity (p = m * v).
Given:
Mass of the bullet (before collision), mb = 0.005 kg
Mass of the block, mblock = 0.5 kg
Final velocity after collision, vfinal = 10 m/s
Let's assume the initial velocity of the bullet before the collision is vbullet.
From the momentum conservation equation, we can write:
(mb * vbullet) + (mblock * 0) = (mb + mblock) * vfinal
Since the bullet hits the stationary block, the initial velocity of the block is 0 m/s.
Now we can solve the equation for vbullet:
(0.005 kg * vbullet) + (0.5 kg * 0) = (0.005 kg + 0.5 kg) * 10 m/s
0.005 kg * vbullet = 5 kg * 10 m/s
0.005 kg * vbullet = 50 kg * m/s
vbullet = (50 kg * m/s) / 0.005 kg
vbullet = 10000 m/s
Therefore, the original velocity of the bullet before the collision was 10000 m/s.
1. work = force * distance = 24 N * 1.5 m = ? Joules
... what box?
2. momentum is conserved
v * .005 = 10 * .505