Work
Take g as 9.8 ms-2 or 9.8 Nkg-1
How much work is done if you push a shopping trolley with a constant force of 60N and it moves 5m in a direction parallel to the force?
Take g as 9.8 ms-2 or 9.8 Nkg-1
How much work is done if you push a shopping trolley with a constant force of 60N and it moves 5m in a direction parallel to the force?
W = Fx
W = (60N) (5m)
W = 300 J
A delivery driver lifts a mass of 6.5kg onto the back of a lorry 1.5m from the ground. How much work is done in this energy transfer?
A delivery driver lifts a mass of 6.5kg onto the back of a lorry 1.5m from the ground. How much work is done in this energy transfer?
W = Fx F = (6.5kg) (9.81 Nkg-1) x = 1.5 m
W = (6.5kg) (9.81 Nkg-1) (1.5 m)
W = 95.55 J
How much work is done lifting a 5kg bag 1.2m and place it on the table. In your calculation you assume all the work is done against what?
How much work is done lifting a 5kg bag 1.2m and place it on the table. In your calculation you assume all the work is done against what?
W = Fx F= (5kg) (9.81 Nkg-1) x = 1.2 m
W = (5kg) (9.81 Nkg-1) (1.2m)
W = (5kg) (9.81 Nkg-1) (1.2m)
W = 58.8 J
Work is done against gravity
How much work is done pulling a bag of rubbish 10m across a field by pulling on a string at 400 to the horizontal with a force of 250N?
W = F cos θ x
W = (250 N) (cos 40) (10 m)
W = 1915 J
A toy car has a mass of 110g and its clockwork engine exerts a force of 0.12N. Unfortunately it is not well made and its wheels are at a 130 angle to its direction of motion. What work does it do in travelling 25cm along a heavily carpeted floor?
A toy car has a mass of 110g and its clockwork engine exerts a force of 0.12N. Unfortunately it is not well made and its wheels are at a 130 angle to its direction of motion. What work does it do in travelling 25cm along a heavily carpeted floor?
W = F cos θ x x = 25cm = 0.25m
W = (0.12N) (cos 13) (0.25 m)
W = 0.029J
A pyramid builder is organising his gang of acolytes to pull a large stone block up a ramp. The stone weighs 2.5t and the ramp has a height of 7m. The ramp is 150m long and the acolytes exert a force of 1.2 kN. How much work to they do? How much work is done on the stone to lift it through a height of 7m? How much work is done against friction?
Work done lifting stone = weight of stone x vertical height
W = Fx F = (2.5 x 103 kg) (9.81 Nkg-1) x = 7m
W = (2.5 x 103 kg) (9.81 Nkg-1) (7m)
W = 1.72 x 105 J
Work along ramp = Tension in rope x length of slope
W= Fx F= 1.25 kN = 1.2 x 103 N x =150m
W = (1.25 x 103 N) (150m)
W = 1.88 x 105 J
Work done against friction= work done pulling stone up ramp – work done against gravity
W against friction = (1.88 x 105 J ) – (1.72 x 105 J) = 16000 J
An 80kg baseball player slides to a halt from a speed of 8 ms-1 in a distance of 4m. What is the average stopping force exerted on him by the ground? How much work is done on him? Where does the energy come from?
We need to calculate the force exerted on the baseball player by the ground. For this we need to know his acceleration.
Using v2 = u2 + 2as
Rearranging 2as = v2 – u2
Cross multiplying a = (v2 – u2) / 2s v= 0 u = 8 ms-1 s = 4m
a = (0 - 82) / 8m = 64/8 = 8ms-2
Using Newton’s second law F=ma = (80 kg) (8ms-2) = 640 N
W= Fx F= 640N x = 4m
W = (640 N) (4m)
W = 2560 J
KE of player
A 50g ball bearing is dropped from a height of 50cm into a tray of fine sand and embeds itself to a depth of 1.5cm. What was the average vertical force exerted on the ball bearing by the sand? How much work is done on the ball bearing?
A 50g ball bearing is dropped from a height of 50cm into a tray of fine sand and embeds itself to a depth of 1.5cm. What was the average vertical force exerted on the ball bearing by the sand? How much work is done on the ball bearing?
Ignoring frictional losses
Work done on sand = Work done lifting ball to height of 50cm
Work done lifting ball = Fx F = (0.05kg) (9.81 Nkg-1) x = 0.5m
W = Fx = (0.05kg) (9.81 Nkg-1) (0.5m) = 0.25 J
Work done on sand = average force exerted by sand x depth of ball
W= Fx
F= W/x = 0.25J / 0.015m = 16.4 N
