Question

If an ice block of mass 42kg moves on a rough surface (${\mu }_k=0.1$). If the initial velocity of block is 4 $m{s}^{-1}$, then the amout of ice melted as a result of friction before the block comes to rest in gm is

• A. $0.5$
• B. $1$
• C. $80$
• D. $16$

Answer 1

Answer: (B)

$1$

$m_1$ is the mass of ice melted
$L$ is the latent heat of fusion
$\frac{m{v}^2}{2}=m_1\times L$
$\therefore m_1=\frac{42\times 4^2}{2\times 0.336\times {10}^6}={10}^{-3}kg=1gm$