Your cousin Throckmorton skateboards from rest down a curved, frictionless ramp. If we treat Throcky and his skateboard as a particle, he moves through a quarter-circle with radius R = 3.00 m. Throcky and his skateboard have a total mass of 25.0 kg. (a) Find his speed at the bottom of the ramp. (b) Find the normal force that acts on him at the bottom of the curve.
The Correct Answer and Explanation is:
To solve this physics problem, we apply principles from conservation of mechanical energy and Newton’s second law in circular motion.
(a) Find his speed at the bottom of the ramp
Concept:
Since the ramp is frictionless, mechanical energy is conserved. The potential energy at the top transforms entirely into kinetic energy at the bottom.
Step 1: Gravitational Potential Energy at Top
At the top of the quarter-circle ramp, Throcky has gravitational potential energy given by:PE=mghPE = mghPE=mgh
where:
- m=25.0 kgm = 25.0 \, \text{kg}m=25.0kg
- g=9.80 m/s2g = 9.80 \, \text{m/s}^2g=9.80m/s2
- h=R=3.00 mh = R = 3.00 \, \text{m}h=R=3.00m
PE=25.0×9.80×3.00=735 JPE = 25.0 \times 9.80 \times 3.00 = 735 \, \text{J}PE=25.0×9.80×3.00=735J
Step 2: Kinetic Energy at Bottom
At the bottom, all this energy becomes kinetic:KE=12mv2KE = \frac{1}{2}mv^2KE=21mv2
Set PE=KEPE = KEPE=KE:735=12×25.0×v2735 = \frac{1}{2} \times 25.0 \times v^2735=21×25.0×v2735=12.5v2⇒v2=73512.5=58.8735 = 12.5v^2 \quad \Rightarrow \quad v^2 = \frac{735}{12.5} = 58.8735=12.5v2⇒v2=12.5735=58.8v=58.8=7.67 m/sv = \sqrt{58.8} = 7.67 \, \text{m/s}v=58.8=7.67m/s
✅ Answer (a):
Throcky’s speed at the bottom is 7.67 m/s
(b) Find the normal force at the bottom of the curve
Concept:
At the bottom, Throcky is moving in a circular path and experiences centripetal acceleration directed upward, provided by the net force. The normal force and his weight both act here:
- Weight W=mgW = mgW=mg (downward)
- Normal force NNN (upward)
The net upward force equals centripetal force:N−mg=mv2RN – mg = \frac{mv^2}{R}N−mg=Rmv2
Solve for NNN:N=mg+mv2RN = mg + \frac{mv^2}{R}N=mg+Rmv2N=25.0×9.80+25.0×(7.67)23.00N = 25.0 \times 9.80 + \frac{25.0 \times (7.67)^2}{3.00}N=25.0×9.80+3.0025.0×(7.67)2N=245+25.0×58.83.00=245+14703.00=245+490=735 NN = 245 + \frac{25.0 \times 58.8}{3.00} = 245 + \frac{1470}{3.00} = 245 + 490 = 735 \, \text{N}N=245+3.0025.0×58.8=245+3.001470=245+490=735N
✅ Answer (b):
The normal force at the bottom is 735 N
Explanation (300 words):
In this problem, Throckmorton begins at rest atop a frictionless, curved ramp shaped like a quarter circle with radius R=3.00 mR = 3.00 \, \text{m}R=3.00m. As there’s no friction, the mechanical energy of the system is conserved. This means his initial gravitational potential energy transforms entirely into kinetic energy at the bottom.
At the top, the only form of energy is gravitational potential energy PE=mghPE = mghPE=mgh, where hhh is the vertical distance fallen, equal to the radius of the quarter circle. By substituting the given values into the energy equation, we find that the total potential energy is 735 J. Since all of this converts to kinetic energy at the bottom, we use the kinetic energy formula KE=12mv2KE = \frac{1}{2}mv^2KE=21mv2 and solve for vvv. This gives a speed of 7.67 m/s.
To find the normal force at the bottom, we consider the fact that Throcky is moving in a circular path and requires centripetal force to maintain that motion. The normal force from the ramp surface must not only support his weight but also provide this extra force to keep him in a curved path. This net force requirement is given by N−mg=mv2RN – mg = \frac{mv^2}{R}N−mg=Rmv2. Solving this gives a total normal force of 735 N at the bottom, which is greater than his static weight due to the added effect of circular motion.
Thus, energy conservation and circular dynamics together explain the motion and the forces at play.
