Betty weighs 427 N and she is sitting on
a playground swing seat that hangs 0.22 m above the ground. Tom pulls the swing back and releases it when the seat is 0.88 m above the ground.
The acceleration of gravity is 9.8 m/s2.
Betty is moving 3.5967 m/s when the swing passes through its lowest position.
If Betty moves through the lowest point at 0.6 m/s, what is the magnitude of the work done on the swing by friction?
Answer in units of J
To find the magnitude of the work done on the swing by friction, we need to calculate the change in kinetic energy.
The work done by friction can be obtained using the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy.
The kinetic energy of an object can be calculated using the formula:
KE = 0.5 * m * v^2
where m is the mass of the object and v is its velocity.
In this case, Betty weighs 427 N, which can be converted to kilograms by dividing by the acceleration due to gravity (9.8 m/s^2):
m = 427 N / 9.8 m/s^2 = 43.57 kg
Given that Betty moves through the lowest point at 0.6 m/s, the initial velocity (u) is 3.5967 m/s.
Using the formula for kinetic energy, we can find the initial kinetic energy (KE_initial) as:
KE_initial = 0.5 * m * u^2
KE_initial = 0.5 * 43.57 kg * (3.5967 m/s)^2
Solving this equation gives us the initial kinetic energy.
Next, we need to find the final kinetic energy (KE_final) when Betty is moving at 0.6 m/s.
KE_final = 0.5 * m * v^2
KE_final = 0.5 * 43.57 kg * (0.6 m/s)^2
Now, we can find the change in kinetic energy:
ΔKE = KE_final - KE_initial
Finally, the magnitude of the work done on the swing by friction is equal to the change in kinetic energy (ΔKE).
Therefore, the answer will be the same as the magnitude of the change in kinetic energy (ΔKE) in units of joules (J).