Find the Ramp Angle for a 3kg Block on a Frictionless Ramp

AI Thread Summary
To find the angle of a frictionless ramp for a 3kg block accelerating at 3.8 m/s², the relevant physics equations involve the relationship between acceleration, gravitational force, and the angle of the ramp. The block's acceleration indicates a specific angle, which can be calculated using trigonometric functions. The possible angle options provided are 12.1°, 16.3°, 22.8°, 25.2°, and 37.4°. The discussion encourages participants to apply their knowledge of physics equations to solve for the angle. Engaging with the problem can lead to a clearer understanding of the forces at play on the ramp.
vincentp07
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A 3kg block is released from rest at the top of a frictionless ramp 1.5m long. The acceleration of the blick is measured to be 3.8m/s/s. What is the angle x of the ramp?

all answers in degrees

a)x= 12.1
b)x=16.3
c)x=22.8
d)x=25.2
e)x=37.4
 
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Hey vincentp07,
Where is your attempt at this qs?
 
Welcome to PF!

vincentp07 said:
A 3kg block is released from rest at the top of a frictionless ramp 1.5m long. The acceleration of the blick is measured to be 3.8m/s/s. What is the angle x of the ramp?

Hi vincentp07! Welcome to PF! :smile:

What equations do you think might be useful in this case? :smile:
 
Thread 'Variable mass system : water sprayed into a moving container'

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}...
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