A spring with a spring-constant 2.2 N/cm is compressed 28 cm and released. The 4 kg mass skids down the frictional incline of heigh 43 cm and inclined at a 22 degrees angle
(g = 9.8 m/s^2)
The path is frictionless except for a distance of 0.9 m along the incline which has a coefficient of friction of 0.4
What is the final velocity of the mass?
I decided to solve this into 2 parts.
One part is when it's at the top with the spring which I found was conserving energy
so I had (mgh + .5kx^2) = (mgh + .5mv^2) and found the v when the block released which became the initial v in the incline
Then the incline's equation (wasn't conserved) was E = Efinal - Einitial = .5mvfinal^2 - (.5mvinitial^2 + mgh) = Fx (F being the frictional force)
and I solved for v final
I don't see what I'm doing wrong but I keep getting the wrong answer
Thank you!!
OK, I agree with part1.
1/2 mv^2=1/2 kx^2
v= sqrt(k/m .28^2)=6.57m/s
does that match yours?
Now, work done by friction down the plane= mg*mu*cosTheta*distance
= 4*9.8*.4*cos22*.9= 13.1J
so KE at the base of the plane
1/2 mvf^2=initialGPE+initialKE-friction
= 4*9.8*Vf^2=4*9.8*.43+1/2 4*6.56^2-13.1
=39.2 Vf^2=16.9+86.2-13.1=90.0
Vf=1.52m/s
check all that.
Well so to convert k constant to N/m
would it be 22 N/m instead of 2.2 N/cm?
no, and I erred above, it becomes 220N/m (I put 2200 above).
well since there's an x component of the gravity
wouldn't that affect the force of friction?
For my force i got (0 = theta)
mgsin0 - (mu)mgcos0
is that wrong?
The force of friction is mu*forcenormal
mu*mgCosTheta in this case, theta is 22 deg.
it's still wrong :(
i've tried my way and your way and it's still wrong
okay i got it
thank you so much
i had my signs all wrong
thank you again!
hey mimi what was your final answer?
i have a similar problem and im trying to see if my answer is wrong or not.
To find the final velocity of the mass, you are correct in breaking down the problem into two parts: the compression and release of the spring, and the motion down the incline.
Let's first calculate the initial velocity of the mass when it is released from the compressed spring.
1. Compression and Release of the Spring:
Using the equation for potential energy stored in a spring, we have:
Potential Energy = (1/2)kx^2
where k is the spring constant (2.2 N/cm) and x is the compression distance (28 cm).
Converting the spring constant to N/m:
k = 2.2 N/cm = 2.2 N/cm * (1 m / 100 cm) = 0.022 N/m
Converting the compression distance to meters:
x = 28 cm = 28 cm * (1 m / 100 cm) = 0.28 m
Now we can calculate the potential energy:
Potential Energy = (1/2)(0.022 N/m)(0.28 m)^2 = 0.00086224 J
Since there is no loss of energy, this potential energy will be converted entirely into kinetic energy when the mass is released.
Using the equation for kinetic energy:
Kinetic Energy = (1/2)mv^2
where m is the mass (4 kg) and v is the initial velocity.
Rearranging the equation, we can solve for v:
v = sqrt((2 * Kinetic Energy) / m)
Substituting the values:
v = sqrt((2 * 0.00086224 J) / 4 kg) = 0.0376 m/s (initial velocity)
This initial velocity will be the starting point for the motion down the incline.
2. Motion Down the Incline:
You correctly recognized that energy is not conserved due to the presence of friction along the incline.
The gravitational potential energy at the top of the incline is converted into three forms of energy: translational kinetic energy, rotational kinetic energy, and work done by friction.
Using the equation for gravitational potential energy:
Potential Energy = mgh
where m is the mass (4 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the incline (43 cm = 0.43 m).
Potential Energy = (4 kg)(9.8 m/s^2)(0.43 m) = 16.808 J
Now let's calculate the work done by friction along the incline.
The normal force (N) can be calculated as the component of the weight perpendicular to the incline:
N = mg * cos(θ)
where θ is the angle of the incline (22 degrees).
N = (4 kg)(9.8 m/s^2) * cos(22 degrees) = 34.283 N
The frictional force (F) can be calculated as:
F = μN
where μ is the coefficient of friction (0.4).
F = (0.4)(34.283 N) = 13.7132 N
The work done by the frictional force over a distance (d) is given by:
Work = Fd
where d is the distance along the incline (0.9 m).
Work = (13.7132 N)(0.9 m) = 12.3419 J
Now we can calculate the final kinetic energy using the principle of work and energy:
Final Energy = Initial Energy - (Work + Losses)
Final Energy = (1/2)mv^2 - (Potential Energy + Work)
Substituting the values:
(1/2)(4 kg)v^2 = 16.808 J - (12.3419 J)
(1/2)(4 kg)v^2 = 4.4661 J
v^2 = (4.4661 J) / (2 * 4 kg) = 0.55651 J/kg
v = sqrt(0.55651 J/kg) = 0.7453 m/s (final velocity)
Therefore, the final velocity of the mass when it reaches the bottom of the incline is 0.7453 m/s.