Question 8.7: Crate Sliding Down a Ramp A 3.00-kg crate slides down a ramp......
Crate Sliding Down a Ramp
A 3.00-kg crate slides down a ramp. The ramp is 1.00 \mathrm{~m} in length and inclined at an angle of 30.0^{\circ} as shown in Figure 8.10. The crate starts from rest at the top, experiences a constant friction force of magnitude 5.00 \mathrm{~N}, and continues to move a short distance on the horizontal floor after it leaves the ramp.
(A) Use energy methods to determine the speed of the crate at the bottom of the ramp.
(B) How far does the crate slide on the horizontal floor if it continues to experience a friction force of magnitude 5.00 \mathrm{~N} ?
(A) Conceptualize Imagine the crate sliding down the ramp in Figure 8.10.
The larger the friction force, the more slowly the crate will slide.
Categorize We identify the crate, the surface, and the Earth as the system. The system is categorized as isolated with a nonconservative force \operatorname{acting}.
Analyze Because v_{i}=0, the initial kinetic energy of the system when the crate is at the top of the ramp is zero. If the y coordinate is measured from the bottom of the ramp (the final position of the crate, for which we choose the gravitational potential energy of the system to be zero) with the upward direction being positive, then y_{i}=0.500 \mathrm{~m}.
Write the expression for the total mechanical energy of the system when the crate is at the top:
E_{i}=K_{i}+U_{i}=0+U_{i}=m g y_{i}
Write an expression for the final mechanical energy:
E_{f}=K_{f}+U_{f}=\frac{1}{2} m v_{f}^{2}+0
Apply Equation 8.16:
\Delta E_{\mathrm{mech}}=\Delta K+\Delta U=-f_{k}d (8.16)
\Delta E_{\text {mech }}=E_{f}-E_{i}=\frac{1}{2} m v_{f}^{2}-m g y_{i}=-f_{k} d
Solve for v_{f} :
(1) v_{f}=\sqrt{\frac{2}{m}\left(m g y_{i}-f_{k} d\right)}
Substitute numerical values:
v_{f}=\sqrt{\frac{2}{3.00 \mathrm{~kg}}\left[(3.00 \mathrm{~kg})\left(9.80 \mathrm{~m} / \mathrm{s}^{2}\right)(0.500 \mathrm{~m})-(5.00 \mathrm{~N})(1.00 \mathrm{~m})\right]}=2.54 \mathrm{~m} / \mathrm{s}
(B) Analyze This part of the problem is handled in exactly the same way as part (A), but in this case we can consider the mechanical energy of the system to consist only of kinetic energy because the potential energy of the system remains fixed.
Write an expression for the mechanical energy of the system when the crate leaves the bottom of the ramp:
E_{i}=K_{i}=\frac{1}{2} m v_{i}^{2}
Apply Equation 8.16 with E_{f}=0 :
E_{f}-E_{i}=0-\frac{1}{2} m v^{2}=-f_{k} d \rightarrow \frac{1}{2} m v^{2}=f_{k} d
Solve for the distance d and substitute numerical values:
d=\frac{m v^{2}}{2 f_{k}}=\frac{(3.00 \mathrm{~kg})(2.54 \mathrm{~m} / \mathrm{s})^{2}}{2(5.00 \mathrm{~N})}=1.94 \mathrm{~m}
Finalize For comparison, you may want to calculate the speed of the crate at the bottom of the ramp in the case in which the ramp is frictionless. Also notice that the increase in internal energy of the system as the crate slides down the ramp is f_{k} d=(5.00 \mathrm{~N})(1.00 \mathrm{~m})=5.00 \mathrm{~J}. This energy is shared between the crate and the surface, each of which is a bit warmer than before.
Also notice that the distance d the object slides on the horizontal surface is infinite if the surface is frictionless. Is that consistent with your conceptualization of the situation?