A Boy of mass 25 kg climbs atop a playground slide of height h= 3 m. When the boy slides down the inclined board, the work done to overcome friction is 550 Joule. What is the boy’s speed before he touches the ground.
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To find the boy's speed before he touches the ground, we can use the principle of conservation of energy. The initial potential energy at the top of the slide is converted into the final kinetic energy at the bottom of the slide.
The potential energy at the top of the slide is given by the formula:
Potential energy = mass * gravity * height
where,
mass = 25 kg (given)
gravity = 9.8 m/s^2 (acceleration due to gravity)
height = 3 m (given)
Substituting the values, we get:
Potential energy = 25 kg * 9.8 m/s^2 * 3 m = 735 Joule
Now, let's assume the final kinetic energy at the bottom of the slide is K. Since the work done to overcome friction is 550 Joule, the remaining energy must be in the form of kinetic energy.
Hence,
Kinetic energy = Potential energy - Work done to overcome friction
K = 735 Joule - 550 Joule
K = 185 Joule
The formula for kinetic energy is:
Kinetic energy = 0.5 * mass * velocity^2
We know the mass of the boy (25 kg) and the kinetic energy (185 Joule). We can rearrange the formula to solve for velocity:
Velocity = √(2 * Kinetic energy / mass)
Substituting the values, we get:
Velocity = √(2 * 185 Joule / 25 kg)
Velocity = √(370 J / 25 kg)
Velocity = √14.8 m^2/s^2
Velocity ≈ 3.85 m/s
Therefore, the boy's speed before he touches the ground is approximately 3.85 m/s.