A 29 kg child, with a 4.5 kg backpack on his back, first stands on a sidewalk and then jumps up into the air. Find the magnitude and direction of the force on the sidewalk from the child when the child is (a) standing still and (b) in the air. Now find the magnitude and direction of the net force on Earth due to the child when the child is (c) standing still and (d) in the air.
total mass = 33.5kg
total force = 9.8 * 33.5
when standing, that's it.
in the air, no force
on earth, force is the same when standing
in the air, F = GmM/r², so it depends on how high he can jump.
To solve this problem, we need to consider Newton's third law of motion, which states that for every action, there is an equal and opposite reaction.
(a) When the child is standing still:
The child exerts a gravitational force on the sidewalk equal in magnitude to his weight. The force can be calculated as the product of the child's mass (29 kg) and the acceleration due to gravity (9.8 m/s^2). Therefore, the magnitude of the force on the sidewalk is:
Force = 29 kg * 9.8 m/s^2 = 284.2 N
The direction of this force is downward (opposite to the force of gravity acting on the child).
(b) When the child is in the air:
When the child jumps into the air, his feet push against the ground, causing an upward force. According to Newton's third law, the ground exerts an equal and opposite force on the child. Hence, the magnitude and direction of the force on the sidewalk are the same as in part (a) - 284.2 N downward.
(c) When the child is standing still:
The net force on Earth due to the child can be calculated by subtracting the gravitational force on the child from the gravitational force exerted by Earth on the child. The magnitude of the force on Earth is:
Force = (29 kg + 4.5 kg) * 9.8 m/s^2 = 333.05 N
The direction of this force is upward (opposite to the force of gravity acting on the child).
(d) When the child is in the air:
When the child is in the air, the gravitational force exerted on him is still the same. Since the child is no longer in contact with the ground, there is no additional force exerted by the child on Earth. Therefore, the net force on Earth remains the same as in part (c) - 333.05 N upward.
To find the magnitude and direction of the force on the sidewalk from the child, we need to consider Newton's third law of motion, which states that for every action, there is an equal and opposite reaction.
(a) When the child is standing still on the sidewalk, the force exerted by the child downwards (due to gravity) will have an equal and opposite force exerted by the sidewalk upwards. Therefore, the magnitude of the force on the sidewalk is the weight of the child, which is given by the formula:
Force = mass × acceleration due to gravity
The acceleration due to gravity near Earth's surface is approximately 9.8 m/s^2.
Force = 29 kg × 9.8 m/s^2 = 284.2 N
The direction of the force on the sidewalk is in the upward direction.
(b) When the child jumps up into the air, the force on the sidewalk is still equal and opposite to the force exerted by the child downwards. Therefore, the magnitude of the force on the sidewalk remains the same at 284.2 N, and the direction is still in the upward direction.
(c) Now let's consider the net force on Earth due to the child when the child is standing still. The child exerts a downward force due to gravity, and according to Newton's third law, there is an equal and opposite force exerted by the Earth on the child. Therefore, the magnitude of the net force on Earth is also 284.2 N, but the direction is in the downward direction (opposite to the force on the child).
(d) When the child is in the air, the net force on Earth is still due to the force of gravity acting on the child. Since there are no other significant forces acting horizontally or vertically on the child, the magnitude of the net force remains the same at 284.2 N, and the direction is still in the downward direction.