an inclined plane makes an angle of 30.0 degree with the horizontal ground. A 40.0 kg mass is dragged up the ramp to the maximum vertical height of 5.00 m.

1. Determine the work done to the mass if the ramp is frictionless (in J)

2. Using the coservation of energy, determine the average force of friction if 300 kJ of work was done in dragging the mass up the ramp(in N)

3. Dtermine the coefficient of friction

4. Dtermine the percent efficiency of dragging the mass up the ramp in part b/c(in %)

### 1 Answer | Add Yours

1. vertical height =5.0 m

Thus displacemet in inclined plane along ramp say d =5.cosec (30)

=5 (2)=10 m

Thus horizontal work done= mg (inclined displacemet along ramp ) cos(30)

= 40 x 9.8 x 10 x sqrt(3)/2

= 3394.82 J

2. Average Force of friction =work done / dispacement

=`300/d=300/10 =30 N`

3. Norma force N

N= mg cos (theta)

=40 x 9.8 cos(30)

=339.48 N

Thus

`f= mu N`

f - is frictional force, N is normal force and Mu coefficient of friction

`30=u xx 339.48`

`mu=.088`

4.efficiency =`((339.48-30)/339.48 )xx100`

`=91.16`

### 1 Reply | Hide Replies ▲

Hi, could you please explain the first question?thanks

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