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erik [133]
3 years ago
5

Joel has a mass of 50kg he is spending his firs day on skis. His friend is trying to push him across a level patch of snow. Joel

friend is pushing with 60N of force but Joel does not move why?
Physics
1 answer:
rusak2 [61]3 years ago
3 0

Answer:

Because of frictional force acting in the opposite direction.

Explanation:

The force of push on the body of Joel is 60 N. The body doesn't move. This clearly means that there must be a force acting in the opposite direction that has a magnitude equal to 60 N so that the net force acting on the body of Joel is 0 and hence the body doesn't move forward.

This opposite force is the frictional force which acts between the surface of snow and Joel. The frictional force always opposes relative motion.

So, when Joel's friend pushes him, he tends move him forward. Therefore, frictional force acts in the opposite direction to oppose the motion. If the frictional force is strong enough to stop the force of push, the body won't move.

This frictional force acting on a stationary body is a variable force and has a maximum value known as limiting friction. When the force of push exceeds the limiting friction, the body just starts to move. The body won't move till the force pf push is less than or equal to limiting friction.

Thus, Joel' friend push is less than or equal to the limiting friction and therefore, Joel is not moving.

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A 60-kg cheetah reaches a speed of 30 m/s as it chases its prey.What is the kinetic energy of the cheetah?
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Zero, a hypothetical planet, has a mass of 5.3 x 1023 kg, a radius of 3.3 x 106 m, and no atmosphere. A 10 kg space probe is to
Andrej [43]

(a) 3.1\cdot 10^7 J

The total mechanical energy of the space probe must be constant, so we can write:

E_i = E_f\\K_i + U_i = K_f + U_f (1)

where

K_i is the kinetic energy at the surface, when the probe is launched

U_i is the gravitational potential energy at the surface

K_f is the final kinetic energy of the probe

U_i is the final gravitational potential energy

Here we have

K_i = 5.0 \cdot 10^7 J

at the surface, R=3.3\cdot 10^6 m (radius of the planet), M=5.3\cdot 10^{23}kg (mass of the planet) and m=10 kg (mass of the probe), so the initial gravitational potential energy is

U_i=-G\frac{mM}{R}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{3.3\cdot 10^6 m}=-1.07\cdot 10^8 J

At the final point, the distance of the probe from the centre of Zero is

r=4.0\cdot 10^6 m

so the final potential energy is

U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{4.0\cdot 10^6 m}=-8.8\cdot 10^7 J

So now we can use eq.(1) to find the final kinetic energy:

K_f = K_i + U_i - U_f = 5.0\cdot 10^7 J+(-1.07\cdot 10^8 J)-(-8.8\cdot 10^7 J)=3.1\cdot 10^7 J

(b) 6.3\cdot 10^7 J

The probe reaches a maximum distance of

r=8.0\cdot 10^6 m

which means that at that point, the kinetic energy is zero: (the probe speed has become zero):

K_f = 0

At that point, the gravitational potential energy is

U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{8.0\cdot 10^6 m}=-4.4\cdot 10^7 J

So now we can use eq.(1) to find the initial kinetic energy:

K_i = K_f + U_f - U_i = 0+(-4.4\cdot 10^7 J)-(-1.07\cdot 10^8 J)=6.3\cdot 10^7 J

3 0
3 years ago
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