A suitcase of 10 kg is being towed along a rough horizontal floor by a force F,

Question

A suitcase of 10 kg is being towed along a rough horizontal floor by a force F, as shown in the figure below. The magnitude of the force F is 50 N, ? = 30°. The coefficient of kinetic friction between the suitcase and the floor is 0.13. The suitcase starts from rest and is towed for 20m along the floor.

1) Calculate the work done by the normal force.
2) Calculate the work done by the towing force. Keep three digits.
3) Calculate the work done by gravitational force.
4) Calcuate the work done by the kinetic frictional force. Keep three digits.
5) Apply the work-KE theorem to find the final speed of the suitcase. Keep three digits.
6) Apply the alternate form of the work-KE theorem to find the final speed of the suitcase. Keep three digits.

Explanation / Answer

given,

mass m=10 kg

force F=50 N,

theta =30 degrees

co-efficient of kinetic friction uk=0.13

initil speed u=0

distance travelled s=20 m

now,

from the diagram,

normal force N=mg-F*sin(theta)

frictional force fk=uk(N)

fk=uk(mg-F*sin(theta))

the net force acting on the suitcase along horizontal line is,

Fnet=ma=F*cos(theta)-uk(mg-F*sin(theta))

now,

1)

work done by the normal force is zero

w=0 (displacement is along normal force is zero)

2)

work done

W=Fnet*s=ma*s

w=(F*cos(theta)-uk(mg-F*sin(theta)))*s

w=(50*cos(30)-0.13(10*9.8-50*sin(30)))*20

w=676.225 J

3)

work done by the gravitational force is zero

w=0 (displacement is along normal force is zero)

4)

work done by the frictional force

wf=fk*s

wf=uk(mg-F*sin(theta))*s

wf=0.13(10*9.8-50*sin(30))*20

wf=189.8 N

5)

by using work -KE energy theorm,

w=1/2*m*v^2-1/2*m*u^2

676.225 =1/2*10*v^2-0

finel speed v=11.629 m/sec

6)

here force on a suit case is ,

Fnet= ma=(F*cos(theta)-uk(mg-F*sin(theta))

a=(F*cos(theta)-uk(mg-F*sin(theta)))/m

a=(50*cos(30)-0.13(10*9.8-50*sin(30)))/10

a=3.381 m/sec^2

now,

v^2-u^2=2*a*s

v^2-0^2=2*3.381*20

===>

final speed v=11.629 m/sec

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