Question

1) A mover pushes a 5.87 kg box with a 48.1 N constant horizontal force up a 14.9° ramp that has a height of 2.56 m. If the ramp is assumed to be frictionless, find the speed of the box as it reaches the top of the ramp using work and energy.

2) A 1100 kg car is traveling 35.5 m/s when it comes up to the bottom of a 14.0° hill. If the car coasts up the hill and the coefficient of rolling friction is 0.450, how far up the hill will the car travel?

Answer 1

Initial energy of the system = 0
Final energy of the system = potential energy + Kinetic energy
= mgH + (1/2)mV²
where m is mass = 5.87 kg
H is height of ramp = 2.56 m
V is velocity at top
Now we know that
Work done = Change in energy
Work done = Force*displacement
Horizontal displacement = x
tan14.9 = 2.56/x
x = 2.56/tan14.9 = 9.62 m
Work done = 48.1*9.62 = 462.722 J
Work done = Final energy - initial energy
462.722 = mgH +(1/2)mV²
462.722= [(5.87*9.81*2.56)] + [(1/2)*5.87*V²]
462.722 = 147.417 + [(1/2)*5.87*V²]
(1/2)*5.87*V² = 462.722 - 147.417
V = 10.365 m/s
(2)
Initial energy of the system = (1/2)mu²
where u is the initial velocity = 35.5 m/s
Now let us assume it went up to the hill = d
Now the final energy of the car = Potential energy
= mgdSin14
We know that
Work done by friction = Intial energy - Final energy
Work done by friction = Friction force*distance (d)
= u_r*(mgCos14)*d
Therefore
u_r*(mgCos14)*d = (1/2)mu² - mgdSin14
u_r*(mgCos14)*d + mgdSin14 = (1/2)mu²
(u_rCos14 + Sin14)mgd = (1/2)mu²
(u_rCos14 + Sin14)gd = (1/2)u²
(0.45Cos14 +Sin14)*9.81*d = (1/2)*35.5²
d = 94.66 m