In the figure, a 4.2 kg block slides along a track from one level to a higher level after passing through an intermediate valley. The track is frictionless until the block reaches the higher level. There a frictional force stops the block in a distance d. The block's initial speed is vo = 5.1 m/s, the height difference is h = 1.0 m, and μk = 0.632. Find d. d μ=0- h -μk

4. A 70 Kg athlete leaps straight up into the air from a trampoline with an initial speed of 7.0 m/s. Take y = 0 to be the position the athlete leaves the trampoline and ignore air resistance. how shoqes di nwob edhW (1 912 a. What is the maximum height the athlete reaches? Step 1) Draw a picture and list the value of all variables at the athlete's starting position and ending position. .£q912 ni nolicups or otni gulą of been woy z9ulev ri telual 10 11J (S 9912 Step 2) Is energy conserved? Yes = write the conservation of energy equation. No write the equation for non- conserved work. 102 (E 9912 nobil yd snob how s Step 3) Plug in values to the equation from step 2 and solve for maximum height. srij 11 m 0.00 to trigisil s of brucny ont moit siqaq to llut 101 Serler abin Toravste srit gnol wor 1010 smit tot svioz uoy tel lliw tens potten b. Solve this problem with kinematic equations and confirm the answers are the same. 0.000,21 A (8 21 (1 q912 Aitnsbl (S q912 smils sris tol svio2 (8…

A vehicle applies it breaks for 102 ft before departing the roadway. The roadway surface friction factor is 0.50. The vehicle continues to break on an unpaved surface with a fiction factor of 0.30, for another 30 ft. The vehicle then strikes a stationary object with an impact of 20 mph. What was the initial speed of the vehicle? The roadway and unpaved surface are level.

Q 1