Question

A solid block of mass 60.0 kg starts from rest and slides down a rough surface with a length of 10.0 m where the coefficient of friction is uk=0.20. The surface is inclined at 40.0°. After sliding 10 meters down the incline it hits a spring with a spring constant of ks = 550.0 N/m. Draw a freebody Diagram indicating all forces. Write out and solve the force and energy equations. How high is the block above the spring at the start? Solve for the normal force Solve for the kinetic friction on the block Solve for EEx = mgsinë – fk Solve for the acceleration down the incline of the block After the block has slide 10.0 m down the incline What is the change in potential energy due to gravity? What is the change in the internal energy of the system due to friction? What is the kinetic energy of the block? What is the speed of the block? What is the net power of the block? Fv The block hits the block on the rough incline How much is the spring compressed by the block? How much spring potential energy does the spring have at this point? How far will the block go back up the incline from the compression point?

Answer 1

Solution.

Colution o migisine mg cose h=Lsino ng "K,- 550-01 7:10 let L theo be the length foo the block slides com) Given m = 60kg o = 40 MK = 0.20 start How high in block above the apring i.e. h= lsino (0) 10 sin40° Th-G.427m R shown in figure in the normal reaction to block (6) R = mg covo EGOR 9.81x0340° R = 450.89 N (co Kinetic friction fx fr HAR fp = 0.208450.89 de a 90.178 N (d) net force along the inclined fp = mg sino -fa Ex CS Scanned with CamScanner

free Gox9.8! Sin 40-10.648 fre: 288.16 N cey let a be the acceleration along the inclined then a : fr m. a : 288.16 [down the inclined] 802 m/s2 down due to gravity energy ja: 4 After block slide lom in potential APE - mgh = 60*9.816.427 36476.98J APE = (9) Change in internal energy due to friction AE = work done aginnt frictica = fx h go.148 X 10 AE = 901.785 (h) Kinetic energy of block KE = APE – AE 8646.98–901.48 KE 2775.25 (i) Speed of block KE SMV 오 my= 2475.2 = 92.50 VE 9.61 m/s CS Scanned with CamScanner

(4) Net power of block P = Fnet XV Fax V = 288.16x9.818 P=2771.53 Watt friction work done (K) Wathen the block hita the spring the kinetic energy of block convert into potential energy of Apping and # KE = {kx?+ fxxx vo Potential energy of spring where x comprented length of oping due to hitting 2775.2 = {x550xx++ 30.178 af block in x+3.76m aqlQ.89 ? ge? +0.3279% - 10.09 = 0 x=3.016 m I) Potential energy of apping PE = 2 x Kx2 = x 550 x (3.0167² PES 2501.47J (m) Now the synten have potential energy of 2501.475 during the bounce this potential energy work against the friction & weight of block do the Ics

At thin time friction force directed down the inclined as shown in figure ti mglsat Atamasina fal= f + ing sing Px mg coso + mg sino =0.246079.11260340*76079.81% Sin 400 F = 468.52N Let x be the rebounce lintance Work done = fen -668.18 xxl2501.47 a'-5.339 m CS Scanned with CamScanner