A block of steel of mass 2 ,kg slides down a | vedclass

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(B) Given: Mass $m = 2 \,kg$, inclination $	heta = \sin^{-1}(3/5)$, so $\sin 	heta = 3/5$ and $\cos 	heta = 4/5$. Distance $s = 80 \,cm = 0.8 \,m$.

Vedclass · Accepted Answer

$0.0114$

Vedclass · Answer

(B) Given: Mass $m = 2 \,kg$, inclination $	heta = \sin^{-1}(3/5)$, so $\sin 	heta = 3/5$ and $\cos 	heta = 4/5$. Distance $s = 80 \,cm = 0.8 \,m$. Specific heat $c = 420 \,J kg^{-1} K^{-1}$. Acceleration due to gravity $g = 10 \,ms^{-2}$.Since the block slides at a constant speed, the work done by friction equals the loss in potential energy. The work done by friction $W_f = f_k \cdot s$, where $f_k = \mu_k N = \mu_k mg \cos 	heta$. Since the speed is constant, $mg \sin 	heta = f_k$.Thus, the heat generated $Q = W_f = (mg \sin 	heta) \cdot s$.Equating heat generated to heat absorbed: $Q = mc \Delta T$.$mc \Delta T = (mg \sin 	heta) s$.$\Delta T = \frac{g \sin 	heta \cdot s}{c} = \frac{10 	imes (3/5) 	imes 0.8}{420}$.$\Delta T = \frac{10 	imes 0.6 	imes 0.8}{420} = \frac{4.8}{420} \approx 0.0114^{\circ} C$.

Column-$I$	Column-$II$
$(a)$ Combined existence of liquid-gaseous state of substance.	$(i)$ Sublimation curve
$(b)$ Combined existence of solid-gaseous state of substance.	$(ii)$ Fusion curve
	$(iii)$ Vaporization curve

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