

A 7.4kg block starts 1.2m above the ground, at the top of a frictionless ramp. At...
A 2.00-kg block starts sliding from the top of a frictionless incline. At the bottom of the incline, the mass slides along a horizonal surface and collides with a spring compressing it a distance x. The spring will compress 1 meter with an applied force of 400 Newtons A. Calculate the speed of the block at the halfway down the incline? B. Calculate the speed of the block at the bottom of the incline? C. How much work is done...
A
4.00 kg block starts sliding from rest from the top of a
frictionless incline, the mass slides along a horizontal surface
and collides with a spring compressing it a distance x. The spring
will compress 3 meters with an applied force of 300N.
A) Calulate the speed of the block at the halfway down the
incline.
B) Calculate the speed of the block at the bottom of the
incline
C) How much work is done on the block by...
A block is placed on a frictionless ramp at a height of 1.35m above the ground. Starting from rest, the block slides down the ramp. At the bottom of the ramp, the box slides onto a frictionless horizontal track without slowing down. What is the speed of the block on the horizontal track?the angle of the track is 44.3 degrees
•A 7.8 kg block slides down a rough incline. The friction experienced by this block is equal to 15 N. The block starts off 3 m above the ground and slides 5 m down this rough ramp until getting to a frictionless plane. This plane is 4 m long. The block then hits a spring, compressing it. The spring constant is 9100 N/m. How much is the spring compressed?
3. A 4.00kg block (initially at rest) slides down a frictionless ramp which is inclined at 42° above horizontal. After sliding for 1.20m, the block hits a spring with spring constant k = 110 a) Calculate the speed of the block just before it hits the spring. b) Calculate maximum distance the spring compresses.
A 5kg block starts at a height of 10m above the ground and slides 20m down a rough surfaced incline. The work done by the frictional force in sliding down the incline is 240 J. It reaches the bottom of the incline and then enters a frictionless vertical loop of radius 2m. The speed of the block at the bottom of the hoop is 10.0m/s. The block performs one revolution and exits the loop.o sinu or mohe imImtombnes biduo nsotinpie...
1
45 kg is released from rest from the top of a rough ramp, with Mass - coefficient of kinetic friction 0.25 between the block and the incline, of height 3.2 m and length d 5.5 m. At the bottom of the ramp, the mass slides on a horizontal, frictionless surface until it compresses a spring of spring constant k 2. 110 N/m. a. Calculate the speed of the mass at the bottom of the ramp? b. How far does...
An m = 2.0kg block sits next to a compressed spring with spring constant k = 4.5N/m. The spring is compressed by x = 3.0m. The spring is released and pushes the box along a surface with no friction. Once freely sliding the block transitions to a ramp with an angle 35 above the horizontal. On the ramp the block experiences a constant force of friction with a coefficient of kinetic friction of μ = 0.15. How far along the ramp’s surface does the block...
A block is placed on a frictionless ramp at a height hof 14.5 m above the ground. Starting from rest, the block slides down the ramp. At the bottom of the ramp, the block slides onto a frictionless horizontal track without slowing down. At the end of the horizontal track, the block slides smoothly onto a second frictionless ramp. 0 = 48.3° By = 25.5 How far along the second ramp does the block travel before coming to a momentary...
A block (6 kg) starts from rest and slides down a frictionless ramp #1 of height 6 m. The block then slides a horizontal distance of 1 m on a rough surface with kinetic coefficient of friction μk = 0.5. Next, it slides back up another frictionless ramp #2. Find the following numerical energy values: 1.Initial gravitational potential energy on Ramp #1: U1G = J 2.Kinetic energy at bottom of Ramp #1 before traveling across the rough surface: K =...