A smooth inclined plane is 6.00m long. The top end of the plane is 1.50m above the bottom end of the plane. A box which has a mass of 35.7kg is pulled up the plane by a rope which is parallel to the plane. The force in the rope is 99.6N. The initial speed of the box is 0. Calculate the speed of the of the box when the box is at the top of the plane.
workin-changePE=finalKE
99.6*6-35.7*9.8*1.50=1/2 *35.7*v^2
solve for velocity v at the top.
To find the speed of the box when it reaches the top of the inclined plane, we can use the principle of conservation of mechanical energy.
First, let's find the gravitational potential energy of the box at the bottom of the plane and at the top of the plane.
The gravitational potential energy (PE) is given by the equation: PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height.
At the bottom of the plane:
PE_bottom = m * g * h_bottom
At the top of the plane:
PE_top = m * g * h_top
Since the box starts from rest, its initial kinetic energy (KE) is zero.
At the bottom of the plane:
KE_bottom = 0
At the top of the plane:
KE_top = 1/2 * m * v^2, where v is the speed of the box at the top.
Applying the principle of conservation of mechanical energy, the sum of initial potential energy and the initial kinetic energy is equal to the sum of final potential energy and the final kinetic energy.
PE_bottom + KE_bottom = PE_top + KE_top
Since KE_bottom and PE_bottom are both zero, the equation simplifies to:
0 = PE_top + KE_top
Substituting the values, we get:
0 = m * g * h_top + 1/2 * m * v^2
Now, let's plug in the known values:
m = 35.7 kg (mass of the box)
g = 9.8 m/s^2 (acceleration due to gravity)
h_top = 1.5 m (height of the top end of the plane above the bottom end)
v = ? (the speed we are trying to find)
0 = (35.7 kg) * (9.8 m/s^2) * (1.5 m) + 1/2 * (35.7 kg) * v^2
Simplifying the equation:
0 = 52.3635 + 17.8255 * v^2
Rearranging the equation:
v^2 = -52.3635 / 17.8255
Taking the square root of both sides:
v = √(-52.3635 / 17.8255)
Since we are calculating the speed, the square root gives us two possible solutions, positive and negative. However, since the box is moving in the positive direction up the inclined plane, we discard the negative solution.
v ≈ 1.703 m/s
So, the speed of the box when it reaches the top of the inclined plane is approximately 1.703 m/s.
To calculate the speed of the box when it reaches the top of the plane, we can use the principle of conservation of energy.
The initial potential energy (PE) of the box at the bottom of the plane is given by the formula: PE = mgh, where m is the mass of the box (35.7 kg), g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height of the bottom end of the plane (1.50 m). Plugging in these values, we get:
PE = (35.7 kg) * (9.8 m/s^2) * (1.50 m) = 514.17 J
Since the initial speed of the box is 0, its initial kinetic energy (KE) is also 0.
According to the principle of conservation of energy, the total mechanical energy (ME) of the box (which is the sum of its potential and kinetic energies) remains constant throughout the motion. Therefore, the total mechanical energy at the bottom of the plane (ME1) should be equal to the total mechanical energy at the top of the plane (ME2).
ME1 = PE + KE = 514.17 J + 0 J = 514.17 J
At the top of the plane, the box will have reached a height of 0, so its potential energy (PE2) will be 0.
ME2 = PE2 + KE2 = 0 + KE2
Since the total mechanical energy remains constant, we can equate ME1 and ME2:
ME1 = ME2
514.17 J = 0 + KE2
KE2 = 514.17 J
The final kinetic energy at the top of the plane (KE2) is given by the formula: KE = (1/2) * mv^2, where m is the mass of the box (35.7 kg) and v is the desired velocity.
Plugging in the values, we get:
514.17 J = (1/2) * (35.7 kg) * v^2
2 * 514.17 J = (35.7 kg) * v^2
1028.34 J = (35.7 kg) * v^2
To find the velocity (v), we can rearrange the equation:
v^2 = (1028.34 J) / (35.7 kg)
v = sqrt((1028.34 J) / (35.7 kg))
Using a calculator, we can evaluate the right-hand side of the equation:
v ≈ sqrt(28.809 J/kg) ≈ 5.36 m/s
Therefore, the speed of the box when it reaches the top of the plane is approximately 5.36 m/s.