Solving for Displacement on a 25° Incline w/ Kinetic Energy

AI Thread Summary
The discussion centers on calculating the distance a sled, carrying a monkey, travels up a 25° incline after being given an initial speed of 3.5 m/s. The combined mass is 17.0 kg, and the coefficient of kinetic friction is 0.20. Participants suggest using kinematic equations and energy conservation principles, emphasizing the need to account for gravitational forces and friction when determining acceleration. The correct approach involves calculating the net force, including the frictional force, and applying the appropriate kinematic formula to find displacement. Overall, the conversation highlights the importance of integrating both kinetic energy and the incline's angle in the calculations.
hshphyss
Messages
45
Reaction score
0
Can anyone help me with this?

In a circus performance, a monkey on a sled is given an initial speed of 3.5 m/s up a 25° incline. The combined mass of the monkey and the sled is 17.0 kg, and the coefficient of kinetic friction between the sled and the incline is 0.20. How far up the incline does the sled move?

I tried using the kinematic equations to find out the displacement. But I was left without the time. I know that KE= .5 x mass x velocity^2 but that isn't going to help me find out the displacement. I was also unsure of where to incorporate the incline angle. Thank-you
 
Physics news on Phys.org
hshphyss said:
Can anyone help me with this?
In a circus performance, a monkey on a sled is given an initial speed of 3.5 m/s up a 25° incline. The combined mass of the monkey and the sled is 17.0 kg, and the coefficient of kinetic friction between the sled and the incline is 0.20. How far up the incline does the sled move?

I tried using the kinematic equations to find out the displacement. But I was left without the time. I know that KE= .5 x mass x velocity^2 but that isn't going to help me find out the displacement. I was also unsure of where to incorporate the incline angle. Thank-you
You have the initial velocity... 3.5m/s
friction co-eff... = 0.2
Ff = -0.2FN
You know that Fn is mgcos theta (Draw an FBD)
ma = - 0.2 mgcos 25
a = -0.2gcos25
To find displacement, use the formula:
v^2 = u^2 + 2as
And solve for s...
You have u, initial velocity, and v, final velocity. (= 0)
And a...
 
You can solve this problem using either the kinematic eqns or by energy conservation method.

Using the kinematic eqns, you could have used the eqn,
v² = u² - 2as
where v is the final velocity, u is the initial velocity, a is the deceleration, s is the distance travelled.
 
to the a= -1.78 and the equation would be
n 0^2=3.5^2+2(-1.78)x and it would come out to 3.45 but that answer is not correct when I enter into the answer checker
 
better not forget gravity!

knowing (or wanting to know) distances is a task for Work & Energy.
At the bottom, we have KE, while at the top we have PE_gravity.
The process has friction force (m g cos(theta)) doing negative Work.
Distance along the ramp is related to height uo the ramp by trig.
 
The acceleration is not 1.78 m/s^2. (Pseudo Statistic's answer ignored the component of weight acting down the incline.)
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top