Question

A physics professor is pushed up a ramp inclined upward at an angle 32.0° above the horizontal as she sits in her desk chair, which moves on frictionless rollers. The combined mass of the professor and chair is 86.0 kg . She is pushed a distance 2.30 m along the incline by a group of students who together exert a constant horizontal force (not parallel to the incline) of 599 N . Dr. Moreno's speed at the bottom of the ramp is 1.70 m/s. Find velocity at the top of the ramp.

Answer 1

`v=2.48m/s`

Step-by-step explanation:

From work-energy theorem we got that:

`W=K_(2)-K_(1)=(1)/(2)m(v_(2)^2-v_(1)^2)` (1)

If we see the free body diagram attached we can calculate the work done by the group of students by knowing its definition given by:

`W=F*s`

Where F is Force and s is distance

`W=(Fcos(32)-wsin(32))(s)`

These are the forces at the x-axis that are causing the work done by the students

`W=((599N)cos(32)-(86kg)(9.8m/s)sin(32))(2.3m)`

`W=141.13J`

Now, we can calculate the velocity at the top of the ramp using (1)

`W=(1)/(2)m(v_(2)^2-v_(1)^2)`

`v_(2)=\sqrt{(2W)/(m)+v_(1)^2}=\sqrt{(2(141.13J))/(86kg)+(1.70m/s)^2}=2.48m/s`

Answer 2

v=1.58 m/s

Step-by-step explanation:

N=F*x*Cos(35)

N=599*2.30m*Cos 35

N=1128.54 J

Kinetic E = ½mv²

Initial kinetic energy

E= (86*(1.7)² )/ 2 = 124.27 J

gravitation potential is 0 at the bottom of the ramp

Potential E = mgh

final potential energy

E= 86 * 9.8 * 2.30m* Sin 35

E= 1111.84 J

final: total energy = 1111.84J + 124.27J = 1236.11 J

final kinetic energy = total energy - potential energy

Ek=1236.11 - 1128.54

Ek=107.57 J

E = ½mv²

107.57 = 86 * v² /2

v² = 2.50

v = 1.58 m/s