a block of mass m=10kg is released from rest on a frictionless incline of angle=30°. Theass can be compressed 2.0cn by a force of 200N. The block momentarily stops when it compresses the spring by 5cm.. How far does fge block mobe down fge incline from its rest position to its stopping point. and what is the speedbof the block just as it touches the sprinv
Gravitational force downhill = m g sin 30 = 98/2 = 49 Newtons
moves down slope distance z
compresses spring distance x = 0.05
energy stored in spring at stop = (1/2)(200/0.02)(0.05)^2 = 12.5 Joules
so loss in potential energy = 12.5 Joules
falling distance = z sin 30 = .5 z
m g (.5 z) = 12.5
98 * .5 z = 12.5
z = 0.255 meters = 25.5 cm = total distance to stop
compresses spring .05 m
so hit spring at .255 - .05 = .25 meters slide distance
so fell .125 meters straight down before hitting spring
v = sqrt (2 g h) = sqrt (2 * 9.8 * .125) = 1.56 m/s
check my arithmetic
the energy at max compression is 1/2 k x^2=1/2 *200N/.02m * .05^2=12.5N
Height it goes upward:
12.5=mgh=10*9.8*h or h=12.5/98 m=12.75cm
which is a distance up the ramp of 12.75/sin30=25.5cm
speed of block as it touches spsring; 1/2 mv^2=12.5 solve for v
Oh, never mind the check. We agree :)
To find how far the block moves down the incline from its rest position to its stopping point, we can use the conservation of mechanical energy.
First, let's calculate the change in potential energy as the block moves down the incline. The potential energy (PE) of the block is given by the formula PE = mgh, where m is the mass (10 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the vertical height difference.
Since the block moves down the incline, the vertical height difference is given by h = d * sin(theta), where d is the distance moved down the inclined plane and theta is the angle of inclination (30 degrees).
h = d * sin(theta) = d * sin(30°) = d * 0.5 (approximately)
The potential energy at the stopping point is PE = mgh = 10 kg * 9.8 m/s^2 * (d * 0.5).
Next, we need to consider the work done by the compressed spring, which is equal to the change in potential energy. The spring constant can be calculated using Hooke's Law: F = kx, where F is the force applied (200 N) and x is the compression of the spring (0.05 m). Rearranging the equation, k = F / x.
The work done by the spring is given by W = 1/2 * k * x^2. Since the block momentarily stops when it compresses the spring by 0.05 m, the work done by the spring is equal to the change in potential energy.
1/2 * k * 0.05^2 = 10 kg * 9.8 m/s^2 * (d * 0.5) (equating the work done to the potential energy)
Now, we can solve for d:
0.5 * (200 N / 0.05 m) * 0.05^2 = 10 kg * 9.8 m/s^2 * (d * 0.5)
20 * 0.05 = 10 * 9.8 * d
1 = 4.9 * d
d = 1 / 4.9 = 0.204 m
Therefore, the block moves down the incline by approximately 0.204 meters from its rest position to its stopping point.
To find the speed of the block just as it touches the spring, we can use the conservation of mechanical energy again. At this point, the block has converted all of its potential energy into kinetic energy.
So, the potential energy at the rest position is PE = mgh = 10 kg * 9.8 m/s^2 * d.
The kinetic energy just as it touches the spring is KE = 1/2 * mv^2, where v is the speed we want to find.
Equating the potential energy at the rest position to the kinetic energy just as it touches the spring:
10 kg * 9.8 m/s^2 * d = 1/2 * 10 kg * v^2
9.8 * d = 0.5 * v^2
v^2 = 19.6 * d
v = sqrt(19.6 * d)
v ≈ sqrt(19.6 * 0.05) ≈ sqrt(0.98) ≈ 0.99 m/s
Therefore, the speed of the block just as it touches the spring is approximately 0.99 m/s.