A 50.0-kg boy runs on a horizontal surface at a speed of 10.0 m/s
and jumps onto a cart just before reaching a slope as shown in the figure. The cart is initially
at rest. If the mass of the cart is 150 kg, what is the maximum height h the boy/cart will reach
up the slope assuming the slope has no friction?
To find the maximum height h the boy/cart will reach up the slope, we need to consider the principle of conservation of mechanical energy.
The mechanical energy is conserved in the absence of any external forces, such as friction in this case. There are two forms of mechanical energy: kinetic energy (KE) and potential energy (PE).
Initially, when the boy is running on the horizontal surface, he has kinetic energy given by:
KE1 = (1/2) * m1 * v^2
where m1 is the mass of the boy and v is his speed. Substituting the given values into the equation, we have:
KE1 = (1/2) * 50 kg * (10 m/s)^2 = 2500 J
When the boy jumps onto the cart, their combined mass becomes m1 + m2, where m2 is the mass of the cart. The final kinetic energy just before reaching the slope can be calculated using:
KE2 = (1/2) * (m1 + m2) * v^2
Substituting the given values into the equation, we have:
KE2 = (1/2) * (50 kg + 150 kg) * (10 m/s)^2 = 10000 J
Since there is no friction, this total mechanical energy is converted into potential energy as the cart and boy move up the slope. At the maximum height h, all the kinetic energy is converted into potential energy. Therefore, we can equate the final potential energy to the initial kinetic energy:
PE = mgh, where g is the acceleration due to gravity (9.8 m/s^2).
PE = KE1 - KE2
mgh = KE1 - KE2
4800 * h = 2500 - 10000
4800 * h = -7500
h = -7500 / 4800
h ≈ -1.56 m
Since we're looking for a positive value for the maximum height, we ignore the negative sign. Therefore, the maximum height the boy/cart will reach up the slope is approximately 1.56 meters.
We don't see the diagram, but assume that the cart is at the beginning of an uphill slope of angle θ with the horizontal.
The problem comes to a matter of conservation of energy, since friction is neglected.
Kinetic energy of the running boy
KE= (1/2) mv²=(1/2)50(10^2)=2500 J.
Assume the height reached is h, then
potential energy is
PE=mgh=50*9.81*h
Since energy is (assumed) conserved, we have KE=PE
50(9.81)h=2500
Solve for h.