Conservation of Energy: block on a table

[Fatima Hasan]

Homework Statement

Homework 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

The Attempt at a Solution

A) [/B]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

 

[Orodruin]

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.

- (0.5 k xƒ^2 - 0.5 k xi^2)
= - (0-100)

What numbers have you used here? Please show your entire computation.

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

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?

 

[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

 

[haruspex]

You have A and B right.

(vfy)^2 = (viy)^2 + 2 a d

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

 

[Fatima Hasan]

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.

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 ?

 

[haruspex]

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 ?

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

 

[Fatima Hasan]

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

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 ?

 

[haruspex]

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 ?

Looks good.