Question

1. A 185 g cart is given a quick push up a frictionless incline, giving it an initial velocity of 1.45 m/s. The incline is at an angle of 10.0 to the horizontal. You must show your work to get full credit for this problem. a) What is the cart's initial kinetic energy, i.e. when it leaves the launcher? KE =mv eay @ztty =C85k9)(145s) = .1343 b) What is the cart's change in kinetic energy (Kr-Ko) from the launch point to the point at which it stops? What is the sign of this change? Is KE gained or lost? AKE-m m = 185)) -Cles(145ms) The signis naganueKE iS 10St This change in KE is due to the work done by gravity, so how much work has been done on c) the cart by gravity as it moved up the incline? Ng = AlaKE W 1343 What is the force acting along the incline due to gravity? F-masine F.l8scgt9mls)Sinio 315 S d) Given the work done and force acting, how far up the incline does the cart move before it e) stops, i.e. the distance along the incline (remember W force distance)? W= F.d de426m How much vertical displacement has occurred during its movement up the incline given the distance traveled and the angle of the incline? 1 g) If KE is lost by the cart, where does it go (remember what is acting on the cart)? h) Near the Earth's surface gravity stores energy (gravitational potential energy) as "mgh". Using conservation of energy determine the vertical displacement of the cart, assuming the initial KE determined above 2. A block of mass 0.800 kg is given an initial velocity vA 1.20 m/s to the right and collides with a light spring of force constant k 50.0 N/m as in the figure below. If the surface is frictionless, calculate the maximum compression of the spring after the collision. al E-a Maximum Compracci

Answer 1

AS Given data m a185 mg cas to m Sino ut.45mls e10 9-10 (a) the ct s fhitial kinetic entj CKE mu axo 1s5 x (1.45) - 1945T CR cb) caat s change Kinetic Pot e Point at hhich it stop the launeh ue (- o-o 1945T Sign. negatve

(C) due tu the cek dee. y gaauity Hs change n ke ak done aav change n KE = - 0-1945T . D cdtca acting le ncine - o185 x 9.8 x n10 O.3148 stoping Let distanta Coot movg mg sin lo - 0.1945 O34-8 x 194 5 o 6179m depla ce h = vetical (8) kinatic enagget cooverled to potntial Cnagj

(6) eath S uate Jaauty staes Engy o ngh mgh O-1945 O194-5 O 185x98 O 1013 m blaxk mas O. end oftial elocit VA = 1-a0mis Constant K 50 ON/m -Pen be ke mosimu Lel Cemprusfan 20 m2- O-8x (1.80 O 02304 k 50 O.15am