Question

h 5. A cart rolls with negligible friction down a ramp that is inclined at an 8 = 32' above level ground. It is released from rest at a height h = 52 cm. What we want to do is to figure out how fast the cart will go when it reaches the bottom of the ramp. Let's begin by establishing the equations that model the motion of the cart. Recall that we are dealing here with an example of constantly accelerated motion. Let the x axis point down the ramp and let x = 0 correspond to the starting position of the cart. What are the equations that describe the motion of the cart? [These should incorporate that both the initial position and velocity are zero. Use 'g' to indicate the gravitational acceleration, 'B' to indicate the angle, and t for the time variable.] v(t) = Submit All Answers Last Answer 1: 5.19*t x(t) = m/s Last Answer 2: 2.59*t^2 m/s Unable to understand formula, tries 3/5. Hint: Hint Video

Answer 1

IR ya. Ľ From Newton's and law of martian mgsino =ma singsing. h i. 32° = ODB ..a=gsind. ingeago ng Displacement = eB h -- Sino v² u²+2 as = 0²+2xg sino xh h=52 an 2.52 m. Sino v=r2gh 2x9.88.52 3.19 m/s. mgsind= ma [R = harmal reaction] : Equations ng es o=R