1. ## Energy

A box of mass $\displaystyle 10kg$ is resting on a sloping platform which is inclined at $\displaystyle 20^\circ$ to the horizontal. The coefficient of friction between the box and the platform is $\displaystyle 0.2$. The box is released from rest and slides down the platform. Calculate:

a) the speed of the box after it has been moving for $\displaystyle 3$ seconds.
b) the potential energy lost by the box during this time.
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I've done part (a) but I cannot get my head around energy. Can I have help on part (b) please. Thanks.

2. Hmm, lost in PE = gain in KE + work done against force of friction ?

3. hi,

Part 1)
Weight of the mass, $\displaystyle W = mg = 10\times 9.81 N = 98.1 N$
normal force acting on the mass by the platform, N = $\displaystyle m\times g\times cos(20^o)=98.1\times cos(20^o) = 92.1838 N$
force from the weight that is pulling the mass parallel down the slope $\displaystyle F = m\times g\times sin(20^o) = 98.1\times sin(20^o) = 33.55217N$
frictional force,$\displaystyle F_{r} = \mu \times N = 0.2\times 92.1838 = 18.43676 N$
resultant force pulling the mass down parallel to the slope,$\displaystyle R =F - R_r = 15.11541 N$

from Newton's second law, F = ma
$\displaystyle R = ma$
$\displaystyle F - R_r = 10\times a$
$\displaystyle 15.11541 = 10\times a$
$\displaystyle acceleration, a = \frac{15.11541}{10} = 1.511541 ms^{-2}$

Part 2)
Using principle of conservation of energy,
Gain in kinetic energy + other energy lost(due to friction) = Lost in gravitational potential energy
$\displaystyle \frac{1}{2} mv^2 + \mu N S= mgh$

where,
v = final velocity of mass after 3 sec under acceleration
S = distance moved by the mass during the 3 second acceleration
h = vertical distance moved by the mass

Finding the final velocity of the mass:
using $\displaystyle v = u + at$:
u = 0
a = 1.511451
t = 3

$\displaystyle v = 0\times 3 + 1.511451\times 3$
$\displaystyle v =4.534353 ms^{-1}$

Kinetic energy obtained:
$\displaystyle \frac {1}{2}m\times v^2 = \frac{1}{2} \times 10 \times 4.534353^2$
$\displaystyle \frac {1}{2}m\times v^2 =102.8017856 Joule$

Finding the distance moved parallel to the plane:
using $\displaystyle S =ut+\frac{1}{2}a\times t^2$
u = 0
t = 3
a = 1.511451

$\displaystyle S =ut+\frac{1}{2}a\times t^2$
$\displaystyle S =0\times 3+\frac{1}{2}\times 1.511451\times 3^2$
$\displaystyle S =0\times 3+\frac{1}{2}\times 1.511451\times 3^2$
$\displaystyle S = 6.8015295 m$

Other energy lost(due to friction):
$\displaystyle LostInFriction = \mu \times N \times S$
$\displaystyle LostInFriction = 0.2\times 92.1838 \times 6.8015295$
$\displaystyle LostInFriction = 125.398167Joule$

Substituting into the principle of conservation of energy:
Gain in kinetic energy + other energy lost(due to friction) = Lost in gravitational potential energy
102.8017856 + 125.398167 = Lost in potential energy
Lost in potential energy = 228.2121958 Joule

There is another method to obtain the lost in gravitational potential energy. Much faster but if you are to answer in an exam this method of using the principle of conservation of energy is the best method to use. Good luck

There is another method to obtain the lost in gravitational potential energy.
I must admit that I'm at a loss. What method are you thinking of?

-Dan

5. directly finding the gravitational potential energy:

Lost in gravitational potential energy = mgh

we already found that the mass will travel 6.801529 m down the plane. this corresponds to a drop in $\displaystyle 6.801529\times sin(20^o) = 2.362599m$ vertically

hence from:
Lost in gravitational potential energy = mgh
$\displaystyle = 10 \times 9.81 \times 2.362599$
$\displaystyle =228.2060985 Joules$