A block of a certain mass is placed on a roug | vedclass

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(D) From the velocity-time graph,the acceleration $a$ of the block is the slope of the line:$a = \frac{v - u}{t} = \frac{20 - 0}{3} = \frac{20}{3} 	e

Vedclass · Accepted Answer

$2.2$

Vedclass · Answer

(D) From the velocity-time graph,the acceleration $a$ of the block is the slope of the line:$a = \frac{v - u}{t} = \frac{20 - 0}{3} = \frac{20}{3} 	ext{ m/s}^2$Since the block is moving,the friction acting on it is kinetic friction,$f_k = \mu_k N = \mu_k mg$.Given $\mu_k = 0.25$ and $g = 10 	ext{ m/s}^2$,the kinetic friction is $f_k = 0.25 	imes m 	imes 10 = 2.5m$.Applying Newton's second law of motion,$F - f_k = ma$:$20 - 2.5m = m 	imes \frac{20}{3}$$20 = m \left( \frac{20}{3} + 2.5 ight)$$20 = m \left( \frac{20 + 7.5}{3} ight) = m \left( \frac{27.5}{3} ight)$$m = \frac{20 	imes 3}{27.5} = \frac{60}{27.5} \approx 2.18 	ext{ kg}$.Rounding to the nearest value,the mass is $2.2 	ext{ kg}$.

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