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Naily [24]
3 years ago
6

A father uses a rope to pull a child on a sled at a constant speed. of the ones listed which forces are present?

Physics
1 answer:
Amanda [17]3 years ago
4 0

All the forces may be present.

The child on the sled has a weight- which is due to the force of gravity on the child and the sled.

The sled and the child exert a force on the ground equal to the combined weight of the sled and the child. The ground exerts a normal force on the sled.

The force used by the father to pull the sled is the applied force.

The sled slides on the ground and as a result, the force of friction exists between the ground and the sled, directed opposite to the direction of the motion of the sled.

The father pulls the sled using a rope. As a result, the rope is under Tension.

As the sled moves it also experiences a force of air resistance, which is dependent on the sled's speed.

However, since the father pulls the sled along with constant speed, the sum of all the forces acting on the sled is zero.

Since there is no movement in the upward or downward directions, the weight of the child and the sled is equal to the normal force acted upon the sled by the ground.

The force applied by the father on the rope is equal to the tension in the rope.

Friction and air resistance act opposite to the direction of motion of the sled. for the sled to move at constant speed, the tension in the rope must be equal to the sum of the forces due to friction and air resistance.

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Answer:

v_a=0.8176 m/s

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Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being m_a and m_b the masses of pucks a and b respectively, the initial momentum of the system is

M_1=m_av_a+m_bv_b

Since b is initially at rest

M_1=m_av_a

After the collision and being v'_a and v'_b the respective velocities, the total momentum is

M_2=m_av'_a+m_bv'_b

Both momentums are equal, thus

m_av_a=m_av'_a+m_bv'_b

Solving for v_a

v_a=\frac{m_av'_a+m_bv'_b}{m_a}

v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}

v_a=0.8176 m/s

The initial kinetic energy can be found as (provided puck b is at rest)

K_1=\frac{1}{2}m_av_a^2

K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J

The final kinetic energy is

K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2

K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J

The change of kinetic energy is

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

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