Your cat “Ms.”(mass 9.00 kg) is trying to make it to the top of a frictionless ramp 2.00m long and inclined 19.0 degrees above the horizontal. Since the poor cat can’t get any traction on the ramp, you push her up the entire length of the ramp by exerting a constant 50.0N force parallel to the ramp. If Ms. is moving at 2.00m/s at the bottom of the ramp, what is her speed when she reaches the top of the incline? What is the cat's speed v2 when she reaches the top of the incline?
work added up the ramp: force*distance=50*2=100joules
work extraced by gravity= 9(9.8*2*sin19)
final KE=initial KE+workdone-PEgained=1/2 * 9(9.8)2^2+100-9*(9.8)2*sin19
from final KE, solve for final velocity.
push force up ramp = 50 N
work done by push = 50 * 2 = 100 Joules
that 100 Joules = gain in Potential + gain in kinetic
gain in potential = m g h = 9 * 9.81 * 2 sin 19 = 57.5 J
so
100 = 57.5 +(1/2)9(V^2 - 2^2)
42.5 = 4.5(V^2 -4) = 4.5 V^2 - 18
60.5 /4.5 =V^2
V = 3.67 m/s
V = 0.816 m/s
Well, I must say, your cat seems to be on quite an uphill journey. Let's calculate her speed at the top of the incline, shall we?
First, we need to break down the forces acting on Ms. Along the inclined plane, there are two forces: the force you applied (50.0N) and the gravitational force pulling her down the ramp.
The force of gravity can be calculated using the formula: F_gravity = mass * g, where g is the acceleration due to gravity (approximately 9.8 m/s^2). In this case, F_gravity = 9.00 kg * 9.8 m/s^2 = 88.2 N.
Now, let's resolve the force parallel to the ramp (50.0N) into its x and y components. The component of the force parallel to the ramp is given by F_parallel = force * sin(theta), where theta is the angle of inclination (19.0 degrees in this case). Therefore, F_parallel = 50.0N * sin(19.0) ≈ 16.4N.
Since there are no other forces acting on the cat in the horizontal direction, the net force (F_net) acting on Ms. is given by F_net = F_parallel = 16.4N (since the gravitational force is perpendicular to the motion).
Now, we can use Newton's second law in the horizontal direction, which states that F_net = mass * acceleration. Rearranging the formula, we have acceleration = F_net / mass = 16.4N / 9.00 kg ≈ 1.822 m/s^2.
Finally, we can use the kinematic equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity (2.00 m/s at the bottom), a is the acceleration (1.822 m/s^2), and s is the distance traveled (2.00 m on the ramp).
Plugging in the values, we get v^2 = (2.00 m/s)^2 + 2 * 1.822 m/s^2 * 2.00 m ≈ 11.288. Taking the square root, we find v ≈ 3.359 m/s.
So, when your cat Ms. reaches the top of the incline, she'll be zooming along at approximately 3.359 m/s. Time to give her a round of applause for conquering that frictionless ramp!
To find the cat's speed when she reaches the top of the incline, we can use the principle of conservation of mechanical energy.
First, let's find the change in potential energy (ΔPE) and kinetic energy (ΔKE) as the cat moves up the incline.
ΔPE = m * g * h
where m is the mass of the cat (9.00 kg), g is the acceleration due to gravity (9.8 m/s²), and h is the vertical height of the incline.
Since the ramp is inclined 19.0 degrees above the horizontal, we can calculate the vertical height using the equation:
h = L * sin(θ)
where L is the length of the ramp (2.00 m) and θ is the angle of inclination (19.0 degrees).
Now, let's calculate ΔPE:
h = 2.00 m * sin(19.0°)
h ≈ 0.6338 m
ΔPE = 9.00 kg * 9.8 m/s² * 0.6338 m
ΔPE ≈ 58.943 J
Next, let's find the initial kinetic energy (KE1) and final kinetic energy (KE2) of the cat.
KE1 = 0.5 * m * v1²
where v1 is the initial velocity of the cat at the bottom of the incline (2.00 m/s).
KE1 = 0.5 * 9.00 kg * (2.00 m/s)²
KE1 = 18.00 J
Since the ramp is frictionless, there is no work being done against friction. Therefore, the change in kinetic energy (ΔKE) is equal to zero.
ΔKE = KE2 - KE1
0 = KE2 - 18.00 J
KE2 = 18.00 J
Now, to find the final velocity (v2) of the cat at the top of the incline, we can use the equation:
KE2 = 0.5 * m * v2²
18.00 J = 0.5 * 9.00 kg * v2²
v2² = 4.00 m²/s²
v2 = √(4.00 m²/s²)
v2 = 2.00 m/s
Therefore, the cat's speed (v2) when she reaches the top of the incline is 2.00 m/s.
force of gravity down ramp = m g sin19 = 9 * 9.81 * .3256 = 28.7 N
push force up ramp = 50 N
net force up = 50 - 28.7 = 21.26 N up ramp
work done by push = 21.26 * 2 = 42.5 Joules
that 42.5 Joules = gain in Potential + gain in kinetic
gain in potential = m g h = 9 * 9.81 * 2 sin 19 = 57.5 J
so
42.5 = 57.5 +(1/2)9(V^2 - 2^2)
- 15 = 4.5(V^2 -4) = 4.5 V^2 - 18
3/4.5 = V^2
V = 0.816 m/s
push force up ramp = 50 N
work done by push = 50 * 2 = 100 Joules
that 100 Joules = gain in Potential + gain in kinetic
gain in potential = m g h = 9 * 9.81 * 2 sin 19 = 57.5 J
so
100 = 57.5 +(1/2)9(V^2 - 2^2)
42.5 = 4.5(V^2 -4) = 4.5 V^2 - 18
60.5 /4.5 =V^2
V = 3.67 m/s