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1 answer:
Answer:
Magnitude of static friction force is 70 sin40° = 44.99 N.
No, it is not necessary that it is maximum static friction.
Normal force is equal to 70 cos40° = 53.62 N.
Explanation:
We apply newton law of moton equation along the plane and perpendicular to plane;
Along the plane,
70 sin 40° = ---------------(1)
70 cos 40° = N --------------(2)
= μN -----------------(3)
So, it depends on the value of μ that the friction is maximum or not .
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(a) 0.165 m/s
The total initial momentum of the astronaut+capsule system is zero (assuming they are both at rest, if we use the reference frame of the capsule):
The final total momentum is instead:
where
is the mass of the astronaut
is the velocity of the astronaut
is the mass of the capsule
is the velocity of the capsule
Since the total momentum must be conserved, we have
so
Solving the equation for , we find
(negative direction means opposite to the astronaut)
So, the change in speed of the capsule is 0.165 m/s.
(b) 520.8 N
We can calculate the average force exerted by the capsule on the man by using the impulse theorem, which states that the product between the average force and the time of the collision is equal to the change in momentum of the astronaut:
The change in momentum of the astronaut is
And the duration of the push is
So re-arranging the equation we find the average force exerted by the capsule on the astronaut:
And according to Newton's third law, the astronaut exerts an equal and opposite force on the capsule.
(c) 25.9 J, 390.6 J
The kinetic energy of an object is given by:
where
m is the mass
v is the speed
For the astronaut, m = 125 kg and v = 2.50 m/s, so its kinetic energy is
For the capsule, m = 1900 kg and v = 0.165 m/s, so its kinetic energy is