# Homework Help: Conservation of Energy: block on a table

1. Jan 1, 2018

### Fatima Hasan

1. The problem statement, all variables and given/known data

2. Relevant equations
∑F=ma
W = - ΔUs
ΔUs = 0.5 k (xƒ)^2 - 0.5 k (xi)^2
W = ΔUs + ΔKE
d = viy t + 0.5 * t^2 * a
3. The attempt at a solution
A)
W = - (0.5 k xƒ^2 - 0.5 k xi^2)
= - (0-100)
W = 100 J
B)
W = ΔUs + ΔKE
100 = 0.5 m vf^2 -0 + 0 - 0.5 k xi ^2
100 = 0.5 * 2 * vf^2 - 0.5 * 2000*(0.1+1.5)
vf = 51.58
C)
∑F=ma

m a = - Fk + F spring
a = ( -4+200)/2 = 196 /2 = 98 m/s^2
d = viy t + 0.5 * t^2 * a
0.3 = 0 + 0.5 * 98 * t^2
t = 0.078 s t = - 0.078 s ( unacceptable)
v = R / t
R = 51.58 * 0.078 = 4.04 m

Please check if it's correct or not.
Thank you

2. Jan 1, 2018

### Orodruin

Staff Emeritus
In general, you should write out your reasoning in words and properly define the notation you are using. Not doing so just makes your post difficult to read and decreases your chances of getting meaningful replies.

You seem to be ignoring a crucial point of the problem statement, namely that the table is rough with a given coefficient of friction. Also, what is the 1.5 doing in your expression?

3. Jan 1, 2018

### Fatima Hasan

I tried to solve it again , here's my work
A)
xi = 0.1 m , xf = 0 m
Ug = 0.5 * k * Δx^2
= 0.5 *2000*(0-0.1)^2
= 10 J
B)
W = Fκd cos θ
Fk = FN μκ = mg μκ
W = 2 * 10 * 0.2 * 1.5 * -1
= - 6 J
W = ΔUg + ΔKE + ΔUs
Δ Ug = 0 (because the height doesn't change)
W = 0.5 k ( xf )^2 - 0.5 k ( xi )^2 + 0.5 m (vf)^2 - 0.5 m (vi)^2
-6 = 0.5 * 2000* (0)^2 - 0.5 * 2000( 0.1 )^2 + 0.5 * 2 * ( vf )^2 - 0.5 *2*(0)^2
v f = √ ((-6+10) / (0.5*2))
v f = 2 m/s
C)
(vfy)^2 = (viy)^2 + 2 a d
(2)^2 = 0 + 2 *a * 1.3
a = 4 / (1.3 *2 ) = 1.5 m/s^2
d = viy t + 0.5 t^2 * a
1.3 = 0 + 0.5 t^2 * 1.5
t = √(1.3 / ( 0.5*1.5)) = 1.31 s ( - 1.31 s unacceptable)
vx = R / t
R = vx * t = 1.316 * 2 = 2.63 m /s

4. Jan 2, 2018

### haruspex

You have A and B right.
Is this supposed to be for the vertical direction? The 2 m/s is horizontal, and the vertical acceleration is known.

5. Jan 2, 2018

### Fatima Hasan

So , no need to use this equation .. I can find the answer by multiblying the horizontal velocity ( 2 m/s) by the time ( 1.36 ) to get the horizontal distance . Isn't it ?

6. Jan 2, 2018

### haruspex

You used that (wrong) equation to get the (wrong) 1.316s.

7. Jan 2, 2018

### Fatima Hasan

Since the initial velocity of y component = 0 m/s ( we have only horizontal velocity ) and the vertical distance was given , so I can use this equation :
d = viy + 0.5 * a t^2
1.3 = 0 + 0.5 *10 * t^2 ( assume that downward direction is positive )
so , t = 0.5 s , t = -0.5 s --> invalid value
R = vx * t = 2 * 0.5 = 1 m
Am I right now ?

8. Jan 2, 2018

Looks good.