C an someone solve these problems for me please? The 3.00-kg object in the figur

Question

Can someone solve these problems for me please?

The 3.00-kg object in the figure is released from rest at a hight of 5.00 m on a curved frictionless ramp. At the foot of the ramp is a spring of force constant 400 N/m. The object slides down the ramp and into the spring, compressing it a distace x before comming momentarily to rest. (a) Find x. (b) Describle the motion object (if any) after the block momentarily comes to rest? [Answer: (a) 0.858 m (b) the energy stored in the compressed spring will accelerate the block, launching it back up the incline and the block will retrace its path, rising to a height of 5.00 m.] A 1500-kg roller coaster car starts from rest a hight H = 23.0 m (as the figure shows) above the bottom of a 15.0-m-diameter loop. If friction is negligible, determine the downward force of the rails on the car when the upside-down car is at the top of the loop. [Answer: 16.7 kN] A 0.17-kg baseball is launched from the roof of a building 12 m above the ground. Its initial velocity is 30 m/s at 40 degree above the horizontal. Assume any effects of air resistance are negligible. (a) what is the maximum height above the groung the ball reaches? (b) What is the speed of the ball as it strikes the ground? [Answer: (a) 31.0 m, (b) 34.0m/s] ALSO LOOK AT ALL OF THE HOMEWORK PROBLEMS, QUIZZES, PROBLEM SESSION PROBLEMS AND EXAMPLES COVERED IN CLASS.

Explanation / Answer

All problems are solved by conservation of energy.

Problem 5) Potential energy of the object at height h is stored in the spring.

mgh = 0.5 * k * x^2

3 * 9.8 * 8 = 0.5 * 400 * x^2

solving for x, we get x =0.858m.

Problem 6) At 15m above the ground, the velocity of the body is calculated using conservartion of energy.

m*g*h = 0.5 * m * v^2

Change in height h = 23 - 15 = 8m

g=9.8 m/sec^2

cancelling m on both sides we get,

( 23 -18 ) * 9.8 = 0.5 * V^2

or v^2 = 156.8

Now, Force at the top is mg - mv^2 / r, which is gravitational - centrifugal

Thus, force at top =

1500 * 9.8 - 1500 * 156.8 / 7.5

= 14.7 - 31.36 = -16.66kN

negative sign indicates that the direction of the force is against the gravity. The final answer being 16.66kN upwards onto the surface of the roller coaster.

Problem 7)

a) The projected velocity is 30 m/s at 40 degress to the horizontal.

Vertical velocity is v *sin( 40 ) = 19.28 m/s upwards

By conservation of kinetic and potential energy,

0.5 * m * v^2 = mgh

solving for h = v^2 / 2g = ( 30 sin40 )^2 / 2 = 18.58 m

Since we launched at the top of 12m building, net height attained by body is 18.58 + 12 = 30.58m

b) Again by conservation of energy,

kinetic at top + potential at top = kinetic at ground

0.5 * m * 30^2 + m * 9.8 * 12 = 0.5 * m * v^2

Solving for v = sqrt( 900 + 2 * 12 * 9.8 ) = sqrt( 1135 ) = 33.689 m/s

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.