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# Answer to Question #37261 in Mechanics | Relativity for alzahra

Question #37261
A 282-kg crate is being pushed across a horizontal floor by a force P that makes an angle of 48.1 ° below the horizontal. The coefficient of kinetic friction is 0.288. What should be the magnitude of P, so that the net work done by it and the kinetic frictional force is zero?

http://edugen.wileyplus.com/edugen/courses/crs3976/art/qb/qu/c06/EAT_12259100087480_6454488344216817.gif
1
2013-12-05T09:03:20-0500
Solution:
m=282kg
Theta=48.1 &deg;
k=0.288

Force P has 2 components : vertical ( P_vert) and Horizontal (P_hor)
P_vert=P*sin(Theta)
P_hor=P*cos(Theta)

ForceFriction=k*N
N- reaction force from the surface
N=Weight+P_vert=m*g+P_vert=m*g+P*sin(Theta)
so

ForceFriction=k*(m*g+P*sin(Theta))

WorkFriction=-ForceFriction*Distance (&quot;-&quot; because ForceFriction and Distance are the opposite direction)

WorkP - work of the force P
WorkP=P_hor*Distance (horizontal projection of the &quot;P&quot; times Distance)

now , the sum of the works is zero

WorkFriction+WorkP=0
-ForceFriction*Distance + P_hor*Distance=0
ForceFriction=P_hor
k*(m*g+P*sin(Theta))=P*cos(Theta)
k*m*g+k*P*sin(Theta)=P*cos(Theta)
k*m*g=P*cos(Theta)-k*P*sin(Theta)
k*m*g=P*(cos(Theta)-k*sin(Theta))

P=k*m*g/(cos(Theta)-k*sin(Theta))

cos(Theta)=cos (48.1 &deg;)= 0.668
sin(Theta)=sin(48.1 &deg;)=0.744

P=0.288*282*9.8/(0.668-0.288*0.744)=795.92/0.454=1753N

P=1753N

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