Question

a block with mass 0.7 kg is forced against a horizontal spring, compressing the spring a distance of 0.2m. The spring constant k is 125 N/m. When released, the block moves on a horizontal tabletop for 0.5m and then up a ramp. The coefficient of kinetic friction between the tabletop and block is 0.3, but the ramp is frictionless. To what vertical height does the block go? show work please

Answer 1

First, we find the velocity of block just after it leaves the spring

1/2kx² = 1/2mv²

v = sqrt (kx² / m)

v = 2.6726 m/s

Now, the friction force acts in the opposite direction of motion, so, using newton's second law

F_net = ma

u_kmg = ma

a = -0.3 * 9.8 = -2.94 m/s²

Now, use kinematics to find the velocity of block just before it goes up the ramp

v² - u² = 2ad

v = sqrt (2.6726² + 2 * -2.94 * 0.5)

v = 2.05 m/s

so, now the speed of block just before it goes up the ramp is 2.05 m/s

the final velocity will be zero,

so,

Here, I think some information is missing from the question. we need the angle of incline to solve for vertical distance reached. you probably have missed some key information while copy, pasting the question.

This is how you can solve for it

first you need to find acceleration on incline. use second law again

mgsin = ma

a = gsin,

here is angle of incline

After you have found the acceleration,

you can use kinematics to find the distance moved along the ramp by using

v² - u² = 2ad

here v = 0 , u = 2.05, a = gsin,

Solve for d.

Notice that question asks for vertical height

so,

use trigonometry

sin = h / d

you know the angle, you have found d and you can very well solve for h (vertical height)