Download Workbooklet 2.5 exercises page 14

Booklet 2.5 page 14 exercises
1. The speed of a body of mass 2.5 kg is increased from 4.0 m/s to6.0 m/s. Calculate:
(a) the gain in kinetic energy of the body;
(b) the work-product of the Resultant Force on the body.
(a)  KE = ½ mv2 - ½ mu2
= ½ m(v2 - u2)
= ½ x 2.5 x (36 – 16)
= +25J
(b)
[W is work product or work done or simply Work, not weight force due to gravity]
W
= F.d
= ma x d
= m (v/ t) x d
= m x v x (d/t)
= m x v x v [ note: v: average velocity during journey ]
= 2.5 x -2 x 5
[ note: v is a positive change ; a increase here]
= +25J
(just to remember :
Which is essentially what we did:
W
= KE
Fxd = ½ mv2 - ½ mu2
part (b) = (a)
2. A body of mass 1.4 kg and travelling at 2.0 m/s is brought to rest over a distance of 0.70 m. Calculate:
(a) the loss in kinetic energy of the body;
(b) the work-product of the Resultant Force on the body;
(c) the size of the Resultant Force on the body.
(a) KE = ½ mv2 - ½ mu2
= ½ m(v2 - u2)
= ½ x1.4x (02 - 22)
= -2.8J
(b) W
= F.d
= ma x d
= m (v/ t) x d
= m x v/t x d
= m x v x d/t
= m x v x v
= 1.4 x -2 x 1
= -2.8J
[where v = 0 – 2 = -2m/s and vaverage = (2 + 0)/2 = 1m/s]
(c )
W
= F.d
[Rearrange] F = W/d
= -2.8 / 0.7
= -4 N ( negative sign here indicates that it is a retarding force)
3. A body of mass 4.5 kg falls from rest through a vertical distance of 1.2 m. Taking g = 10 N/kg,
calculate:
(a) the gravitational force on the body;
(b) the work-product of this gravitational force over the fall;
(c) the loss of potential energy of the body due to the fall;
(d) the gain of kinetic energy due to the fall;
(e) the speed of the body at the end of the fall.
(a) Fg
= mg
= 4.5 x 10
= 45N
(b) W
= F.d
= 45 x 1.2
= 54J
(c) PE   KE
 PE = - 54J
[ All converted to motion ; energy is conserved here]
[ or a loss of 54 Joules]
(d) PE   KE
 KE = 54J
[ All converted to motion ; energy is conserved here]
[ or a gain of 54 Joules]
(e) use the KE forumula here and rearrange to solve:
½ mv2 = 54
54x 2
v
m

108
4.5
 24
2 6
 4.9 m/s
4.
A body is hauled up a ramp 5.0 m long against a constantfriction force of 20 N. Calculate:
(a) the work-product of the friction;
(b) the heat energy produced.
(a) W
= Fr x d
= 20 x 5
= 100J
(b) Fr  heat energy [energy is conserved here]
 Heat produced here = 100J
5.
800 J of energy is supplied to a body to haul it up a frictionless ramp.
While it is being hauled up the ramp its kinetic energy increases by 215 J. How much gravitational
potential energy is gained by the body?
Total energy
 800J

PE
= PE +  KE + Heat energy(due to friction, but here Fr = 0)
= PE + 215J
=
585J
6. A body of mass 2.5 kg is projected vertically with an initial speed of 40 m/s.
(a) What is the initial kinetic energy of the body?
(b) What is the kinetic energy of the body when it reaches its maximum height?
(c) How much potential energy does the body gain as it rises from the point of projection to its maximum height?
(d) What is the work-product of the gravitational force on the body as it rises from the point of projection to its
maximum height?
(e) What is the gravitational force on the body? (g = 10 N/kg.)
(f) What is the maximum height reached?
(a) KE
= ½ mv2
= ½ x 2.5x 402
= 2000J
(b) KE  PE ,
and v = 0m/s at the maximum height  KE = 0J
(c) KE  PE , Total transfer of energy PE = 2000J
(d) W = F x d = mgh = 2000J
(e) W
= mg
= 2.5 x 10
= 25N
[ this W is not work product but the Weight force due to gravity]
(f) W = PE = mgh [rearrange to solve]
 h =W/(mg)
= 2000 / 25
= 80m