Answer :
Explanation:
It is given that,
Mass of the block, m = 3.57 kg
It is drawn to a distance of, d = 4.06 m
Horizontal force acting on the rope, F = 7.68 N
Angle above horizontal, [tex]\theta=15^{\circ}[/tex]
(A) The work done by the rope on the block is W. It can be calculated as :
[tex]W=Fd\ cos\theta[/tex]
[tex]W=7.68\times 4.06\ cos(15)[/tex]
W = 30.11 joules
(b) Let [tex]\mu[/tex] is the coefficient of kinetic friction between block and floor. According to the free body diagram of the given problem. At equilibrium,
[tex]Fcos\theta=\mu(mg-Fsin\theta)[/tex]
[tex]\mu=\dfrac{Fcos\theta}{(mg-Fsin\theta)}[/tex]
[tex]\mu=\dfrac{7.68\times cos(15)}{(3.57\times 9.8-7.68\times sin(15))}[/tex]
[tex]\mu=0.22[/tex]
Hence, this is the required solution.
Answer:
Explanation:
Given
mass of block=3.57 kg
distance =4.06 m
Force on rope(F)=7.68 N
inclination \theta =15
cos component of rope will do work on block
[tex]W=F\cdot s[/tex]
[tex]W=F s \cos 15[/tex]
[tex]W=7.68\times 4.06\times \cos 15[/tex]
W=30.11 J
(b)constant speed implies that Force exerted by rope is being balanced by friction in horizontal direction
[tex]F\cos \theta =f_r[/tex]
[tex]f_r=\mu N[/tex]
[tex]f_r=\mu (mg-F\sin \theta )[/tex]
[tex]f_r=\mu 33.006[/tex]
[tex]7.418=\mu \times 33.006[/tex]
[tex]\mu =0.22[/tex]