Question

A block of mass m-5.9 kg is pulled up a -24 incline as in the figure below with a force of magnitude F 33 N (a) Find the acceleration of the block if the incline is frictionless. (Give the magnitude of the acceleration) m/s2 (b) Find the acceleration of the block if the coefficient of kinetic friction between the block and incline is 0.12. (Give the magnitude of the acceleration. m/s2

Answer 1

(a)

Weight of the block acting vertically down is split into its components

Net Force =Mass*acceleration

$F-mgsin24=ma$

$33N-5.9kg*9.81m/s^{2}*sin24=5.9kg*a$

$33-5.9*9.81*sin24=5.9*a$

$9.46=5.9*a$

ANSWER: ${\color{Red} a=1.6m/s^{2}}$

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(b)

Here Fr is the frictional force, which is acting opposite to movement of block

$Fr=\mu N=\mu mgcos24$

so Net Force =Mass*acceleration

$F-mgsin24-Fr=ma$

$F-mgsin24-\mu mgcos24=ma$

$F/m-gsin24-\mu gcos24=a$

$33N/5.9kg-9.81m/s^{2}*sin24-0.12*9.81m/s^{2}*cos24=a$

$33/5.9-9.81*sin24-0.12*9.81*cos24=a$

ANSWER: ${\color{Red} a=0.528m/s^{2}}$

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