A movie stuntman (mass 80.0kg) stands on a window ledge 5.0m above the floor. Grabbing a rope attached to a chandelier, he swings down to grapple with the movie’s villian (mass 70.0 kg), who is standing directly under the chandelier.(assume that the stuntman’s center of mass moves downward 5.0 m. He releases the rope just as he reaches the villian. (a) with what speed do the entwined foes start to slide across the floor?

If the coefficient of kinetic friction of their bodies with the floor is 0.250, how far do they slide?

The question aims to understand newton’s law of motion, the law of conservation, and the equations of kinematics.

Newton’s law of motion states that the acceleration of any object relies on two variables, the mass of the object and the net force acting on the object. The acceleration of any object is directly proportional to the force acting on it and is inversely proportional to the mass of the object.

A principle that does not change and states a certain property in the course of time within an isolated physical system is called conservation law. Its equation is given as:
\[U_i + K_i = U_f + K_f \]
Where the U is the potential energy and K is the kinetic energy.
The science of explaining the motion of objects using diagrams, words, graphs, numbers and equations is described as Kinematics. The aim of studying the kinematics is to design sophisticated mental models that help in describing the motions of physical objects.

Expert Answer

In the question, it is given that:

Stuntman has a mass of $(m_s) \space= \space 80.0kg$.

Movie’s villain has a mass of $(m_v)= \space 80.0kg$.

The distance between the floor and window is $h= \space 5.0m$.

Part a

Before the collision of the stunt man, the initial velocity and the final height is $0$, therefore the $K.E = P.E$.

\[ \dfrac{1}{2}m_sv_2^2 =  m_sgh\]

\[v_2 = \sqrt{2gh}\]

Therefore the speed $(v_2)$ becomes $\sqrt{2gh}$.

Using the law of conservation, the speed after the collision can be calculated as:

\[v_sv_2= (m_s+ m_v) .v_3\]

Making $v_3$ the subject:

\[v_3 = \dfrac{m_s}{m_s+ m_v} v_2\]

Plugging $v_2$ back in:

\[v_3= \dfrac{m_s}{m_s+ m_v} \sqrt{2gh}\]

Plugging the values and solving for $v_3$:

\[ v_3 = \dfrac{80}{80+ 70} \sqrt{2(9.8)(5.0)} \]

\[ v_3 = \dfrac{80}{150}. 9.89 \]

\[v_3 = 5.28 m/s\]

Part b

The coefficient of kinetic friction of their bodies with the floor is $(\mu_k) = 0.250$

Using Newton’s 2nd law:

\[ (m_s + m_v)a = – \mu_k (m_s + m_v)g \]

Acceleration comes out to be:

\[ a = – \mu_kg \]

Using the Kinematics formula:

\[ v_4^2 – v_3^2 = 2a \Delta x \]

\[ \Delta x = \dfrac{v_4^2 – v_3^2}{2a} \]

Inserting the acceleration $a$ and putting final velocity $v_4$ equals $0$:

\[ = \dfrac{0 – (v_3)^2}{ -2 \mu_kg} \]

\[ = \dfrac{(v_3)^2}{2 \mu_kg} \]

\[ = \dfrac{(5.28)^2}{2(0.250)(9.8)} \]

\[\Delta x = 5.49 m\]

Numerical Answer

Part a:  Entwined foes start to slide across the floor with the speed of $5.28 m/s$

Part b: With kinetic friction of 0.250 of their bodies with the floor, the sliding distance is $5.49m$


On the runway, an airplane accelerates at $3.20 m/s^2$ for $32.8s$ until it finally lifts off the ground. Find the distance covered before takeoff.

Given that acceleration $a=3.2 m/s^2$

Time $t=32.8s$

Initial velocity $v_i= 0 m/s$

Distance $d$ can be found as:

\[ d = vi*t + 0.5*a*t^2 \]

\[ d = (0)*(32.8) + 0.5*(3.2)*(32.8)^2 \]

\[d = 1720m\]

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