Question #6e844

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Question #6e844

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Answer 1 · 2016-12-03T14:49:12

Given

$m \to mass of the block = 9.25 k g$ $m-> "mass of the block"=9.25kg$

$F \to horizontal force applied on the block = 55 N$ $F-> "horizontal force applied on the block"=55N$

$μ_{k} \to coefficient of kinetic friction = 0.175$ $mu_k-> "coefficient of kinetic friction"=0.175$

$F_{fric} \to frictional force resisting the block$ $F_"fric"-> "frictional force resisting the block"$
$= μ_{k} m g = 0.175 \times 9.25 \times 9.8 N = 15.86375 N$ $" "" "=mu_kmg=0.175xx9.25xx9.8N=15.86375N$

So net force acting on the block

$F_{n} = F - F_{fric} = (55 - 15.86375) = 39.13625 N$ $F_n=F-F_"fric"=(55-15.86375)=39.13625N$

So acceleration produced

$a = \frac{F_{n}}{m} = \frac{39.13625 N}{9.25 k g} \approx 4.23 m s^{- 2}$ $a=F_n/m=(39.13625N)/(9.25kg)~~4.23ms^-2$

If time taken to cover the distance $s = 3 m$ $s=3m$ be t sec then by using equation of kinematics we can write

$S = u \times t + \frac{1}{2} a t^{2}$ $S=uxxt+1/2at^2$

$\Rightarrow 3 = 0 \times t + \frac{1}{2} \times 4.23 \times t^{2}$ $=>3=0xxt+1/2xx4.23xxt^2$

$t = \sqrt{\frac{6}{4.23}} s = \approx 1.19 s$ $t=sqrt(6/4.23)s=~~1.19s$

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Question #6e844

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