A 50kg boy runs up the steps to the third floor of height 10.5m in 45s.While at the third floor,the boy drpos a 0.25kg ball and strike the ground.
1)calculate the rate of work done
2)by using the work-energy theorem,calculate the speed of the ball just before it strikes the ground
3)calculate the average force applied on the ball if the impulsive force has acted for 4ms.(assume no energy loss)
1. F = M*g = 50 * 9.8 = 490 N. = Wt. of boy.
P = F*d/t = 490 * 10.5/45 = 114.3 J/s = 114.3 Watts. = Work rate.
2. KE = PE.
0.5M*V^2 = M*g*h
0.50*0.25*V^2 = 0.25**9.8*10.5
V = ?.
In the third solution:
F=mv/t
m=0.25 kg and t=4ms=0.004 s.
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correct.
Let's calculate the answers step-by-step:
1) To calculate the rate of work done, we need to find the amount of work done and divide it by the time taken. The work done can be found using the formula:
Work = Force x Distance
The force acting on the boy while he runs up the steps can be calculated using his weight:
Force = mass x gravity
where mass = 50 kg and gravity = 9.8 m/s^2 (approximate value)
Force = 50 kg x 9.8 m/s^2 = 490 N
The distance the boy covers is the height of the third floor, which is 10.5 m.
Work = 490 N x 10.5 m = 5145 J
Now, we can find the rate of work done by dividing the work by the time taken:
Rate of work done = Work / Time
Rate of work done = 5145 J / 45 s
Rate of work done ≈ 114.33 J/s
Therefore, the rate of work done by the boy is approximately 114.33 J/s.
2) According to the work-energy theorem, the work done on an object is equal to the change in its kinetic energy. Since the ball is dropped from rest, the initial kinetic energy is zero. The work done on the ball is the gravitational potential energy of the ball just before it strikes the ground.
Potential Energy = mass x gravity x height
Potential Energy = 0.25 kg x 9.8 m/s^2 x 10.5 m = 25.725 J
This potential energy is converted into the kinetic energy just before the ball strikes the ground:
Kinetic Energy = 25.725 J
Using the equation for kinetic energy, we can find the velocity (speed) of the ball just before it strikes the ground:
Kinetic Energy = 1/2 x mass x velocity^2
25.725 J = 1/2 x 0.25 kg x velocity^2
Solving for velocity:
velocity^2 = (2 x 25.725 J) / 0.25 kg
velocity^2 ≈ 205.8 m^2/s^2
velocity ≈ √(205.8) ≈ 14.34 m/s
Therefore, the speed of the ball just before it strikes the ground is approximately 14.34 m/s.
3) The impulsive force acting on the ball can be calculated using the impulse-momentum theorem:
Impulse = force x time
The impulse acting on the ball is equal to the change in momentum of the ball:
Impulse = Change in momentum
Using the fact that the impulse (force x time) is equal to the change in momentum, we can set up the equation:
Impulse = force x time = Δmomentum
We are given the time for which the impulsive force has acted, which is 4 ms (0.004 s).
Impulse = force x time = Δmomentum
We need to find the change in momentum, which is equal to the final momentum of the ball just before it strikes the ground:
Momentum = mass x velocity
Momentum = 0.25 kg x 14.34 m/s = 3.585 kg⋅m/s
Impulse = Δmomentum = 3.585 kg⋅m/s
Now, we can find the average force applied on the ball by dividing the impulse by the time:
Average Force = Impulse / Time
Average Force = 3.585 kg⋅m/s / 0.004 s
Average Force = 896.25 N
Therefore, the average force applied on the ball is 896.25 N.
To calculate the answers to these questions, we need to understand some key concepts related to work, energy, and force.
First, let's start by understanding the relationship between work and energy.
1) Work is defined as the transfer of energy that occurs when a force is applied to an object, causing it to move. It is calculated using the equation: Work = Force x Distance x cos(theta), where theta is the angle between the force and the direction of movement.
In this scenario, the work done by the boy to reach the third floor is equal to the change in potential energy, as he is going against gravity. The formula to calculate work done against gravity is: Work = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height.
To calculate the rate of work, we need to divide the work done by the time taken:
Rate of work = Work / Time
Given:
Mass of the boy, m = 50 kg
Height, h = 10.5 m
Time taken, t = 45 s
Calculating the work done:
Work = mgh = 50 kg * 9.8 m/s^2 * 10.5 m = 5145 J (Joules)
Calculating the rate of work:
Rate of work = Work / Time = 5145 J / 45 s = 114.33 W (Watts)
Therefore, the rate of work done by the boy is 114.33 Watts.
Moving on to the second question:
2) According to the work-energy theorem, the work done on an object is equal to its change in kinetic energy. In this case, the ball is dropped from the third floor, and the work done on it is equal to its change in potential energy.
Using the law of conservation of energy, we can equate the potential energy at the top (when the ball was at rest) with the kinetic energy just before hitting the ground.
Potential energy at the top = mgh = 0.25 kg * 9.8 m/s^2 * 10.5 m = 25.725 J (Joules)
Therefore, the kinetic energy just before hitting the ground is also 25.725 J.
We can now use the kinetic energy formula to calculate the speed of the ball just before it strikes the ground:
Kinetic energy = (1/2)mv^2, where m is the mass of the ball and v is its velocity/speed.
Rearranging the equation:
v = sqrt(2 * Kinetic energy / m)
Calculating the speed:
v = sqrt(2 * 25.725 J / 0.25 kg) = sqrt(205.8) m/s ≈ 14.35 m/s
Therefore, the speed of the ball just before it strikes the ground is approximately 14.35 m/s.
Moving on to the third question:
3) To calculate the average force applied on the ball, we can use the formula for impulse, which is the change in momentum of an object.
Impulse = Force * Time = Change in momentum
Since no energy loss is assumed, the impulse given to the ball is equal to its change in momentum.
Using the formula for momentum, P = mv, where m is the mass and v is the velocity, and assuming that the initial velocity is zero (as it was dropped):
Momentum = Change in momentum = mv
Given:
Mass of the ball, m = 0.25 kg
Time for which the impulsive force acts, t = 4 ms = 0.004 s
Calculating the impulse:
Change in momentum = mv = 0.25 kg * (Final velocity - 0) = 0.25 kg * Final velocity
Impulse = Force * Time
=> 0.25 kg * Final velocity = Force * 0.004 s
=> Force = (0.25 kg * Final velocity) / 0.004 s
Using the speed calculated in the previous question, Final velocity = 14.35 m/s:
Force = (0.25 kg * 14.35 m/s) / 0.004 s ≈ 896.875 N
Therefore, the average force applied on the ball, assuming no energy loss, is approximately 896.875 Newtons.