Question #220042

a -A box of mass 10 kg is pulled by a mass-less rope with a force of 40 N. The rope

makes an angle of 30deg with the horizontal. Draw the free-body diagram showing

all forces on the box. Determine the acceleration of the box if the coefficient of

kinetic friction between the floor and the box is 0.2. Take g =10 ms-2 .

b-A ball of mass 100 g strikes a wall with the speed 2.0 ms-1 and bounces off the

wall in the opposite direction with the same speed. Calculate the impulse of the

ball.


1
Expert's answer
2021-07-25T09:42:06-0400

a.

Given:m=10kgF=40Nμ=0.2θ=30°NW=mayNW=0(ay=0)N=WN=mgN=10kg×9.81ms2N=98.1NFR=μ×NFR=0.2×98.1NFR=19.62NFcosθFR=max40N×cos(30°)19.62N=10kg×aa=1.5ms1\textsf{Given}:\\ m=10 kg\\ F=40N\\ \mu=0.2\\ \theta=30\degree\\ \hspace{0.01cm}\\ N-W = ma_y\\ N-W=0 (a_y=0)\\ N=W\\ N=mg\\ N=10kg \times 9.81ms^{-2}\\ N=98.1N\\ \hspace{0.1cm}\\ F_R=\mu\times N\\ F_R=0.2\times98.1N\\ F_R=19.62N\\ \hspace{0.1cm}\\ Fcos\theta-F_R=ma_x\\ 40N\times cos(30\degree)-19.62N=10kg\times a\\ a=1.5ms^{-1}


b.

Given:m=100g=0.1kgu=2.0ms1v=2.0ms1I=ΔPI=m(vu)I=0.1(2.0(2.0)I=0.1×4.0I=0.4NsThe negative value of the impulse indicates that it’sdirection is opposite to the initial direction of the ballbefore impact.\textsf{Given:}\\ m=100g=0.1kg\\ u=2.0ms^{-1}\\ v=-2.0ms^{-1} \\ \hspace{0.1cm} \\ I=\Delta P\\ I=m(v-u)\\ I=0.1(-2.0-(2.0)\\ I=0.1\times-4.0\\ I=-0.4Ns\\ \textsf{The negative value of the impulse indicates that it's}\\ \textsf{direction is opposite to the initial direction of the ball}\\ \textsf{before impact.}


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