A body of mass 40kg is being dragged along the floor by a rope inclined at 60 degrees to the horizontal. The frictional force between the box and the floor is 100N and the tension on the rope is 300N. How much work is done by dragging the box for a distance of 4m
680J
400J
200J
100J
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This is the solution.
The force is supposed to be inclined to the horizontal
So, u will us F = cos60 × tension.
When u get F, subtract the frictional force from it and use work done = Force × distance
m=40kg
Theta=60°
Frictional force=100N
Tension(T or F)= 300N
Distance= 4m
First you should resolve the tension to its horizontal component
Fx=Tcostheta
Fx( horizontal component)=300*cos60°
Fx=150N
Remember that the frictional force has to be overcome before it can move, so we subtract
F=150-100
F=50N (this is the force that actually moved the body)
Work done =force*distance
W=50*4=200J
Workdone = FxDxCos0
W.D=100x4xCos60
W.D=400xCos60
Recall; Cos60=1/2
W.D=400x1/2
W.D=400/2
W.D=200J
Tension= 300N
Frictional force = 100
Distance= 4m
Cos 60° = 0.5
0.5 x tension
= 0.5 x 300 = 150N
Then, 150N - Frictional force
= 150 - 100 = 50N
Work done= Force x distance
= 50 x 4
= 200J
F - FR = ma
F = Tcos60∘ = 300cos60∘ = 150N
FR = 100N, m = 40kg,
ma = 150 - 100 =50N
workdone = F × d = 50 × 4 = 200J. that is the solving.
mass[m]=40
angle[0] ie theta=60
force[f]=100
tension[t]=300
distance[d]=4
formula for work done= f x d
but wherever we see tension and we see an angle , we use [t x cos o - f ] x d
cos 60= 0.5
[300 x 0.5 - 100] x 4
=[150 - 100] x 4
=[50] x 4
=200
Friction has to be overcome for the mass to move.
The weight of the box has been taken into consideration for we to have tension. Therefore Total force (F) = TCos60 - Friction
F= 300Cos60 -100 = 50N
Work =FS
W=50*4=200N
is it only me but when i choose a or any other option in a question when i check the correction it will be changed and i missed it y is it likedis
in question like this please avoid mass. is just to confuse you since they have given u the frictional force so oo there is no need of converting
The selected answer is wrong:
the total useful force is 200n ..by the formula FDcos¶ ..d answer is 400j
