Question

Two carts of masses 200 kg and 300 kg on horizontal rails are pushed apart. Suppose the coefficient of friction between the carts and the rails are same. If the 200 kg cart travels a distance of 36 m and stops, then the distance travelled by the cart weighing 300 kg is

Answer 1

Answer:

24 m

Step by step explanation:

From the laws of motion: $V_f^2=2aS$ (initial velocity is zero).

If we multiply both sides by 'm' we get:

$v^2m=2aSm\\ \\S=\frac{\frac{1}{2}mv^2}{ma}\\E_k=\frac{1}{2}mv^2$ and F = ma.

Therefore, $\frac{S_1}{S_2}=\frac{E_{k1}}{F_1}*\frac{F_2}{E_{k2}}$

In this problem, both acceleration and kinetic energy are constant.

Hence,

$\frac{S_1}{S_2}=\frac{m_2}{m_1} \\\frac{36}{S_2}=\frac{300}{200}\\S_2=\frac{2*36}{3}=24\:m$

Answer 2

The body finally comes to rest. Therefore:

u^2 = 2aS  

where u is initial velocity (Final velocity is zero).  

From the above equation:  

S = u^2 / 2a  

Multiplying Numerator and Denomenator by m, we have  

S = u^2 x m/ 2a x m  

S = Ek / ma  

where Ek = 1/2 x mu^2 and ma = Force.  

In your case Kinetic Energy and Acceleration must be constant. If this is the case, then:  

S1/S2 = Ek1/ F1 x F2 / Ek2  

But Kinetic energy is constant:  

S1/S2 = F2 / F1  

S1/S2 = m2 a / m1 a  

(a = acceleration = constant)  

S1/S2 = m2 / m1  

36/S2 = 300 / 200  

36/S2 = 3 / 2  

S2 = 24 m