A Physics professor is pushed up on a ramp inclined at an angle of 35.0° above the horizontal as he sits in his desk chair that slides on frictionless rollers. The combined mass of the professor and the chair is 86.0 kg. He pushed at a distance 2.95 m along the inclined by a group of students who are exerting a constant force of 591N parallel to the ramp. The speed at the bottom of the ramp is 2.40 m/s. Find the speed at the top of the ramp.
K.E. @ top = K.E. @ bottom + work pushing up - P.E. at top
1/2 m vt^2 = 1/2 m vb^2 + f d - m g h
1/2 * 86.0 * v^2 = ...
... (1/2 * 86.0 * 2.40^2) + (591 * 2.95) - [86.0 * 9.8 * 2.95 * sin(35.0º)]
To find the speed at the top of the ramp, we can use the conservation of mechanical energy. The mechanical energy of the system (professor and chair) is conserved if the only forces acting on the system are conservative forces, such as gravity.
First, we need to find the gravitational potential energy at the top and the bottom of the ramp. The gravitational potential energy (U) is given by the formula:
U = mgh,
where m is the mass of the system, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height.
At the bottom of the ramp, the height is zero, so the gravitational potential energy is also zero.
Next, we need to find the height at the top of the ramp. The top of the ramp and the bottom of the ramp are at the same horizontal level, so the height is also zero at the top.
Since the gravitational potential energy is the same at the top and the bottom of the ramp, it means that the kinetic energy (KE) at the top and the bottom must be equal.
The kinetic energy is given by the formula:
KE = (1/2)mv^2,
where m is the mass of the system and v is the speed.
Let's denote the speed at the top of the ramp as v_top and the speed at the bottom as v_bottom.
Since the speed at the bottom is given as 2.40 m/s, we have:
KE_bottom = (1/2)mv_bottom^2.
At the top of the ramp, the speed is not given, so we express the kinetic energy as:
KE_top = (1/2)mv_top^2.
Since the mechanical energy is conserved, we can equate the kinetic energy at the top and the bottom:
KE_top = KE_bottom.
Therefore:
(1/2)mv_top^2 = (1/2)mv_bottom^2.
Now we can plug in the values:
(1/2)(86.0 kg)(v_top)^2 = (1/2)(86.0 kg)(2.40 m/s)^2.
Simplifying the equation:
v_top^2 = (2.40 m/s)^2.
Taking the square root of both sides:
v_top = 2.40 m/s.
Therefore, the speed at the top of the ramp is 2.40 m/s.