Question #192321

1.A 2kg block is moved by a slope inclined at angle of 37° by a horizontal force of 45N.the block moves a distance of 2m up the plane.if µ=0.2 find

A.the work done by the applied force

B.the work done by gravity

C.the work done by friction

D.the change in kinetic energy of the block

2.A ball of mass 3 kg starts from rest at point p and travels along the track 4m.find the velocity at points Q and R if

A.the track is frictionless

B.the friction force is 20N


1
Expert's answer
2021-05-12T08:37:28-0400

1.A.


W1=Fscos37°=45(2sin37°)cos37°=119.43 (J)W_1=Fs\cos37°=45\cdot(\frac{2}{\sin37°})\cdot\cos37°=119.43\ (J) . Answer


1.B.


W2=Fscos180°=mg2cos180°=29.82 cos180°=39.2 (J)W_2=Fs\cos180°=mg\cdot 2\cdot\cos180°=2\cdot9.8\cdot2\ \cdot \cos180°=-39.2\ (J) . Answer


1.C.


W3=Ffscos180°=μNscos180°W_3=F_fs\cos180°=\mu Ns\cos180°


N=mgcos37+Fsin37°=29.8cos37+45sin37°=28.65 (N)N=mg\cos37+F\sin37°=2\cdot 9.8\cdot \cos37+45\cdot\sin37°=28.65\ (N)


W3=μNscos180°=0.228.652sin37°cos180°=19.04 (J)W_3=\mu Ns\cos180°=0.2\cdot28.65\cdot\frac{2}{\sin37°}\cdot\cos180°=-19.04\ (J) . Answer


1.D.


ΔKE=W1W2W3=119.4339.219.04=61.2 (J)\Delta KE=W_1-W_2-W_3=119.43-39.2-19.04=61.2\ (J) . Answer


2.A.


Assume that the angle of inclination of the track is α=50°\alpha=50°


F=maa=F/m=mgsin50°/m=gsin50°=9.8sin50°=7.5 (m/s2)F=ma\to a=F/m=mg\sin50°/m=g\sin50°=9.8\cdot\sin50°=7.5\ (m/s^2)


v=2sa=247.5=7.75 (m/s)v=\sqrt{2sa}=\sqrt{2\cdot 4\cdot 7.5}=7.75\ (m/s) . Answer


2.B.


F=maa=(FFf)/m=(mgsin50°20)/3=0.84 (m/s2)F=ma\to a=(F-F_f)/m=(mg\sin50°-20)/3=0.84\ (m/s^2)


v=2sa=240.84=2.6 (m/s)v=\sqrt{2sa}=\sqrt{2\cdot 4\cdot 0.84}=2.6\ (m/s) . Answer








Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS