Pad 6:57 PM 50% . \" a usi38ok.theexpertta.com The Expert TA I Human-like Gradin

Question

Pad 6:57 PM 50% . " a usi38ok.theexpertta.com The Expert TA I Human-like Grading, Automated! (6%) Problem 14: A block of mass 46 kg is sitting on a frictionless ramp with a spring at the bottom that has a spring constant of 495 N/m (refer to the figure). The angle of the ramp with respect to the horizontal is 25° Chegg Study Guided Solutions and Study Help | Chegg.com t Status %,ncorrect Answer The provided answer is not correct and re for view no specific feedbackis available Please view available hints and review the relevant material. Close Status Completed Completed Completed Completed Completed Completed Completed Completed Completed Completed Completed Completed Completed Partial Otheexpertta.com S8 33% Part (a) The block, starting from rest, slides down the ramp a distance 65 cm before hitting the spring. How far, in centimeters, is the spring compressed as the block comes to momentary rest? Grade Summary Deductions Potential Ar- 51.1 0% 100% tan0 acoso Submissions Attempts remaining: 6 (0%, per attempt) detailed view sin) cos 789HOME cotan) asin) atanacotan) sinh coshO tanhcotanh0 1 END Degrees O Radians BACKSPACE DELI! CLEAR Submit Hint I give up! Hints: 2% deduction per hint. Hints remaining: 3 Feedback: 2% deduction per feedback 33% Part (b) After the block comes to rest, the spring pushes the block back up the ramp. How fast, in meters per second, is the block moving right after it comes off the spring? -là 33% Part (c) What is the change of the gravitational potential energy in Joules, between the original position of the block at the top of the ramp and the position of the block when the spring is fully compressed?

Explanation / Answer

Part (a) -

Downward acceleration of the block, a = g*sin25 = 9.81*sin25 = 4.14 m/s^2

Velocity of the block after travelling a distance of s = 0.65 m -

use the expression -

v^2 = u^2 + 2*a*s

put the values -

v^2 = 0 + 2*4.14*0.65

=> v = 2.32 m/s

now, suppose x is the compression of the spring.

apply energy conservation -

(1/2)*k*x^2 = (1/2)*m*v*2

=> 0.5*495*x^2 = 0.5*4.6*2.32^2

=> x = 0.2236 m = 22.36 cm

Part (b) -

Velocity of the block after it comes off the spring = velocity of the block before striking the spring = 2.32 m/s

Part (c) -

Vertical height, the block descended, h = (s+x)*sin25 = (0.65 + 0.2236) * sin25 = 0.8736*0.423 = 0.369 m

So, change in the gravitational potential energy of the block = m*g*h = 4.6*9.81*0.369 = 16.66 J

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

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