Question

A physics professor is pushed up a ramp inclined upward at an angle 33.0° above the horizontal as he sits in his desk chair that slides on frictionless rollers. The combined mass of the professor and chair is 90.0 kg. He is pushed a distance 2.00 m along the incline by a group of students who together exert a constant horizontal force of 600 N. The professor's speed at the bottom of the ramp is 2.30 m/s.
A) Find velocity at the top of the ramp.

Answer 1

Answer:

2.51 m/s

Explanation:

Parameters given:

Angle, A = 33°

Mass, m = 90kg

Inclined distance, D = 2m

Force, F = 600N

Initial speed, u = 2.3m/s

From the relationship between work and kinetic energy, we know that:

Work done = change in kinetic energy

W = 0.5m(v² - u²)

We also know that work done is tẹ product of force and distance, hence, net work done will be the sum of the total work done by the force from the students and gravity.

Hence,

W = F*D*cosA - w*D*sinA

w = m*9.8 = weight

=> W = 600*2*cos33 - 90*9.8*2*sin33

W = 45.7J

=> 45.7 = 0.5*m*(v² - u²)

45.7 = 0.5*90*(v² - 2.3²)

45.7 = 45(v² - 5.29)

=> v² - 5.29 = 1.016

v² = 6.306

v = 2.51 m/s

The final velocity is 2.51 m/s

Answer 2

Answer: v = 3.26 m/s

Explanation:

Force applied (F) = 900 N,

mass of professor and chair (m) = 90kg,

g = acceleration due to gravity = 9.8 m/s²,

Fr = frictional force = 0 ( since from the question, the rollers the chair and professor are moving is frictionless),

θ = angle of inclination = 33°

a = acceleration of object =?

From newton's second law of motion, we have that

F - (mg sinθ +Fr) = ma

Where mg sinθ is the horizontal component of the weight of the mass of professor and chair due to the inclination of the ramp.

But Fr = 0

Hence, we have that

F - mg sinθ = ma

600 - (90×9.8×sin30) = 90 (a)

600 - 480.371= 90a

119.629 = 90a

a = 119.629/ 90

a = 1.33 m/s².

But the body started the morning (at the bottom of the ramp) with a velocity of 2.30 m/s²

Hence u = initial velocity = 2.30 m/s², a = acceleration = 1.33m/s², v = final velocity =?, s = distance covered = 2m

By using equation of motion for a constant acceleration, we have that

v² = u² + 2as

v ² = (2.3)² +2(1.33)×(2)

v² = 5.29 + 5.32

v² = 10.61

v = √10.61

v = 3.26 m/s