explaining how Newton’s Three Laws of Motion relate to the marshmallow catapult. AND DONT JUST PUT ANYTHING OR I WILL ERASE YOUR
1 answer:
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1) 4 forces
2) 165 N
3) 225 N
Explanation:
1)
There are in total 4 forces acting on the bicylist:
- The gravitational force on the byciclist, acting vertically downward, of magnitude , where m is the mass of the bicyclist and g is the acceleration due to gravity
- The normal force exerted by the floor on the bicyclist and the bike, N, vertically upward, and of same magnitude as the gravitational force
- The force of push F, acting horizontally forward, given by the push exerted by the bicylist on the pedals
- The air drag, R, of magnitude R = 60.0 N, acting horizontally backward, in the direction opposite to the motion of the bicyclist
2)
The magnitude of the net force on the bicyclist can be calculated by considering separately the two directions.
- Along the vertical direction, we have the gravitational force (downward) and the normal force (upward); these two forces are equal in magnitude, since the acceleration of the bicyclist along this direction is zero, therefore the net force in this direction is zero.
- Along the horizontal direction, the two forces (forward force of push and air drag) are balanced, since the acceleration is non-zero, so we can use Newton's second law of motion to find the net force on the bicylist:
where
is the net force
m = 75.0 kg is the mass of the bicyclist
is its acceleration
Solving, we find the net force:
3)
In this part, we basically want to find the forward force of push, F.
We can rewrite the net force acting on the bicyclist as
where:
F is the forward force of push
R is the air drag
We know that:
is the net force on the bicyclist
R = 60.0 N is the magnitude of the air drag
Therefore, by re-arranging the equation, we can find the force generated by the bicylicst by pedaling:
Answer:19.5 m
Explanation:
Given
coefficient of kinetic Friction
Initial speed
Friction is present so it tries to stop to the object and stops it completely after moving certain distance let say s
maximum deceleration provided by friction is
using equation of motion
where