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

Answer:

v=1.58 m/s

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

Answer 2

Answer:

$v=2.48m/s$

Explanation:

From work-energy theorem we got that:

$W=K_{2}-K_{1}=\frac{1}{2}m(v_{2}^2-v_{1}^2)$ (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)

$W=\frac{1}{2}m(v_{2}^2-v_{1}^2)$

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