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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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4 years ago
A hot air balloon is hovering in the air when it drops a 40 Kg food package to some lost golfers. If the package is dropped from
UNO [17]
We can calculate this with the law of conservation of energy. Here we have a food package with a mass m=40 kg, that is in the height h=500 m and all of it's energy is potential. When it is dropped, it's potential energy gets converted into kinetic energy. So we can say that its kinetic and potential energy are equal, because we are neglecting air resistance:

Ek=Ep, where Ek=(1/2)*m*v² and Ep=m*g*h, where m is the mass of the body, g=9.81 m/s² and h is the height of the body.

(1/2)*m*v²=m*g*h, masses cancel out and we get:

(1/2)*v²=g*h, and we multiply by 2 both sides of the equation

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5 0
3 years ago
A 10 kg monkey climbs up a massless rope that runs over a frictionless tree limb and back down to a 15 kg package on the ground.
pshichka [43]

Answer:

A. 4,9 m/s2

B. 2,0 m/s2

C. 120 N

Explanation:

In the image, 1 is going to represent the monkey and 2 is going to be the package.  Let a_mín be the minimum acceleration that the monkey should have in the upward direction, so the package is barely lifted. Apply Newton’s second law of motion:

\sum F_y=m_1*a_m_i_n = T-m_1*g

If the package is barely lifted, that means that T=m_2*g; then:

\sum F_y =m_1*a_m_i_n=m_2*g-m_1*g

Solving the equation for a_mín, we have:

a_m_i_n=((m_2-m_1)/m_1)*g = ((15kg-10kg)/10kg)*9,8 m/s^2 =4,9 m/s^2

Once the monkey stops its climb and holds onto the rope, we set the equation of Newton’s second law as it follows:

For the monkey: \sum F_y = m_1*a \rightarrow T-m_1*g=m_1*a

For the package: \sum F_y = m_2*a \rightarrow m_2*g - T = m_2*a

The acceleration a is the same for both monkey and package, but have opposite directions, this means that when the monkey accelerates upwards, the package does it downwards and vice versa. Therefore, the acceleration a on the equation for the package is negative; however, if we invert the signs on the sum of forces, it has the same effect. To be clearer:

For the package: \sum F_y = -m_2*a \rightarrow T-m2*g=-m_2*a \rightarrow m_2*g -T=m_2 *a

We have two unknowns and two equations, so we can proceed. We can match both tensions and have:

m_1*a+m_1*g=m_2*g-m_2*a

Solving a, we have

(m_1+m_2)*a =(m_2 - m1)*g\\\\a=((m_2-m_1)/(m_1+m_2))*g \rightarrow a=((15kg-10kg)/(10kg+15kg))*9,8 m/s^2\\\\a= 2,0 m/s^2

We can then replace this value of a in one for the sums of force and find the tension T:

T = m_1*a+m_1*g \rightarrow T=m_1*(a+g)\\\\T = 10kg*(2,0 m/s^2+9,8 m/s^2) \\\\T = 120 N

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3 years ago
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