Tyson throws a shot put ball weighing 7.26 kg. At a height of 2.1 m above the ground, the mechanical energy of the ball is 172.1
2 answers:
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Answer:
A)
B)
C)
Explanation:
This question is incomplete. The full question was:
<em>A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then jumps on his skateboard and descends down the ramp. His speed at the bottom of the ramp is vf = 6.2 m/s. </em>
<em>Part (a) Write an expression for the work, Wf, done by the friction force between the ramp and the skateboarder in terms of the variables given in the problem statement. </em>
<em>Part (b) The ramp makes an angle θ with the ground, where θ = 30°. Write an expression for the magnitude of the friction force, fr, between the ramp and the skateboarder. </em>
<em>Part (c) When the skateboarder reaches the bottom of the ramp, he continues moving with the speed vf onto a flat surface covered with grass. The friction between the grass and the skateboarder brings him to a complete stop after 5.00 m. Calculate the magnitude of the friction force, Fgrass in newtons, between the skateboarder and the grass.</em>
For part A), we make a balance of energy to calculate the work done by the friction force:
For part B), we use our previous value for the work:
Solving for friction force:
For part C), we first calculate the acceleration by kinematics and then calculate the module of friction force by dynamics:
Solving for a:
Now, by dynamics:
Refer to the diagram shown below.
Because the ramp is slippery, ignore dynamic friction.
Let m = the mass of the frog.
g = 9.8 m/s²
The KE (kinetic energy) at the bottom of the ramp is
KE₁ = (1/2)*(m kg)*(5 m/s)² = 12.5 m J
Let v = the velocity at the top of the ramp.
The KE at the top of the ramp is
KE₂ = (1/2)*m*v²= 0.5 mv² J
The PE (potential energy) at the top of the ramp relative to the bottom is
PE₂ = (m kg)*(9.8 m/s²)*(1 m) = 9.8m J
Conservation of energy requires that
KE₁ = KE₂ + PE₂
12.5m = 0.5mv² + 9.8m
0.5v² = 2.7
v = 2.324 m/s
Answer: 2.324 m/s
Because the ramp is slippery, ignore dynamic friction.
Let m = the mass of the frog.
g = 9.8 m/s²
The KE (kinetic energy) at the bottom of the ramp is
KE₁ = (1/2)*(m kg)*(5 m/s)² = 12.5 m J
Let v = the velocity at the top of the ramp.
The KE at the top of the ramp is
KE₂ = (1/2)*m*v²= 0.5 mv² J
The PE (potential energy) at the top of the ramp relative to the bottom is
PE₂ = (m kg)*(9.8 m/s²)*(1 m) = 9.8m J
Conservation of energy requires that
KE₁ = KE₂ + PE₂
12.5m = 0.5mv² + 9.8m
0.5v² = 2.7
v = 2.324 m/s
Answer: 2.324 m/s