A 16-kg child sits on a 5-kg sled and slides down a 143-meter, 30-degree slope, to the nearest m/s what is his or her speed at the bottom?
They want you to neglect friction and use conservation of energy. The masses will not affect the result.
g*L*sin30 = V^2/2
L = 143 m
g = 9.81 m/s^2
Solve for V, and round to nearest m/s
The m/s number that you get will correspond to well over 100 miles per hour - a very unsafe value.
There will actually be appreciable friction unless they are on an ice-covered luge run, and using no brakes or foot dragging.
This is a rather unrealistic problem.
THANKYOU
Thank you drwls, there is another question that I am not getting really well right now. It is this: Two skaters are in the exact center of a circular frozen pond. Skater 1 pushes skater 2 off with a force of 100 N for 1.4 seconds. If skater 1 has a mass of 30 kg and skater 2 has a mass of 74 kg, what is the relative velocity (v1 - v2) after the push to the nearest hundredth of a m/s? After reaching the other shore, how fast, to the nearest tenth of a m/s, must skater 1 run around the lake to meet skater 2 at the opposite shore?
Thanks
To calculate the speed of the child at the bottom of the slope, we can use the principles of physics, specifically the concept of conservation of energy. Here's how you can solve it step by step:
1. Start by determining the potential energy (PE) of the child and sled at the top of the slope. The formula for potential energy is PE = m * g * h, where m is the mass, g is the acceleration due to gravity (9.8 m/s²), and h is the height.
PE_top = (mass_child + mass_sled) * g * h_top
Given:
mass_child = 16 kg
mass_sled = 5 kg
h_top = 0 (since the child is at the top of the slope)
Substitute these values:
PE_top = (16 kg + 5 kg) * 9.8 m/s² * 0 m
PE_top = 21 kg * 9.8 m/s² * 0 m
PE_top = 0 J (Joules)
2. Next, calculate the kinetic energy (KE) of the child and sled at the bottom of the slope. The formula for kinetic energy is KE = (1/2) * m * v², where m is the mass and v is the velocity.
KE_bottom = (1/2) * (mass_child + mass_sled) * v_bottom²
Given:
mass_child = 16 kg
mass_sled = 5 kg
v_bottom = ? (this is what we want to find)
Substitute these values and solve for v_bottom:
KE_bottom = (1/2) * (16 kg + 5 kg) * v_bottom²
KE_bottom = 10.5 kg * v_bottom²
3. Since we have assumed no energy losses due to friction or other factors, the potential energy at the top should be equal to the kinetic energy at the bottom.
PE_top = KE_bottom
Substitute the values:
0 J = 10.5 kg * v_bottom²
4. Rearrange the equation to solve for v_bottom:
v_bottom² = 0 J / 10.5 kg
v_bottom² = 0
v_bottom ≈ 0 m/s
Therefore, the speed of the child at the bottom of the slope is approximately 0 m/s, indicating that there is no motion or the child has come to a complete stop. This result suggests that there might be some energy losses due to friction or other factors that have affected the motion of the sled and child.