Question

A roller coaster of mass $2000\text{kg}$ has a speed of $v_1=10\text{m/s}$ as it goes over the top of a $15\text{m}$ high hill. Then it goes over another hill of height $10\text{m}$ and at the crest of that hill, the roller coaster is moving with speed $v_2=13\text{m/s}$. How much is the work done by friction on the roller coaster during its movement between the hills ? Take $g=10{\text{m/s}}^2$.

• A. $15500\text{J}$
• B. $-15500\text{J}$
• C. $-31000\text{J}$
• D. $31000\text{J}$

Answer 1

The correct option is C $-31000\text{J}$

Applying work energy theorem on the roller coaster from $A$ to $B$
$W_g+W_f+W_N=\Delta KE.....\left(i\right)$


Now work done by gravity from $A$ to $B$ is:
$W_g=+mg(H_A-H_B)=2000\times 10\times (15-10)=100000\text{J}$

Work done by the normal throughout the journey is $\text{zero}$, since normal is always $\perp$ to the velocity of roller coaster at each point, hence $W_N=\overrightarrow{N}.\overrightarrow{dS}=0$

Work done by friction ($W_f$) will be negative as it will always act opposite to the velocity of roller coaster to prevent slipping.

Now, $\Delta KE=\frac{1}{2}m\left(v_2^2-v_1^2\right)$
$=\frac{1}{2}\times 2000\times \left(169-100\right)=69000\text{J}$

Putting all values in Eq $\left(i\right)$:
$100000+W_f+0=69000$
$\therefore W_f=69000-100000=-31000\text{J}$