Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Classical central-force problem wikipedia , lookup
Relativistic mechanics wikipedia , lookup
Density of states wikipedia , lookup
Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup
Heat transfer physics wikipedia , lookup
Work (physics) wikipedia , lookup
Internal energy wikipedia , lookup
Hunting oscillation wikipedia , lookup
Kinetic energy wikipedia , lookup
Physics 6A Ch. 7 – In Class Worksheet - Solutions 1) To pass a slow-moving truck, you want your 1300 kg car to speed up from 13.4 m/s to 17.9m/s in 3.0s. How much work is done on the car during this acceleration? What is the minimum power required for this pass? 2 2 W = ∆K = 1 m v 2f − v i2 = 1 (1300kg) 17.9 m − 13.4 m 2 2 s s W = 91,553J ( Power = ) ( ) ( ) Work 91553 J = = 30,518 Watts Time 3 sec 2) An 82.0kg mountain climber is in the final stage of the ascent of 4301-m-high Pikes peak. What is the change in gravitational potential energy as the climber gains the last 100.0m of altitude? We have a formula for potential energy: ∆Ugrav = mg(∆y ) Setting y=0 at the initial height of the climber, we have yi=0m and yf=100m Plugging these values in we get: ∆Ugrav = (82kg) 9.8 m2 (100m − 0m) = 80,360J s 3) A 1.70kg blocks slides on a horizontal, frictionless surface until it encounters a spring with a force constant of 955 N/m. The block compresses the spring by 4.60cm before turning around and sliding in the opposite direction. Find the initial speed of the block. (Ignore air resistance and any energy lost when the block initially contacts the spring) Use Conservation of Energy. Initial energy is all v initial kinetic, final energy is all elastic potential energy. E f = Ei 4.6cm V=0 www.clas.ucsb.edu/staff/vince final 1 kx 2 = 1 mv 2 2 2 955 N (0.046m)2 = (1.7kg)v 2 m ⇒ v = 1 .1 m s ( ) Physics 6A Ch. 7 – In Class Worksheet (Page 2) - Solutions 4) Deep in the forest, a 17.0g leaf falls from a tree and drops straight to the ground. If its initial height was 5.30m and its speed on landing was 1.3m/s, how much non-conservative work was done on the leaf? We will use conservation of energy. Initially the leaf has gravitational potential energy, but no kinetic energy (not moving). When it hits the ground it has kinetic energy. If we define y=0 at the ground, then Ugrav=0 when the leaf hits. So we have the following values: Initial: yi=5.3m ; vi=0 Final: yf=0 ; vf=1.3 m/s 5.3m Our formula is: Ef = Ei + Wother (we need to find W other in this problem) 2 ½ mvf = mgyi + W other 2 2 ½ (0.017kg)(1.3 m/s) = (0.017kg)(9.8 m/s )(5.3m) + Wother Wother = -0.869 J (negative work because friction slows the leaf down) 5) When a force of 120.0N is applied to a certain spring, it causes a stretch of 2.25cm. What is the potential energy of this spring when it compressed by 3.50cm? First we need to find the spring constant k. We use the force formula Fspring=kx 120N = (k)(0.0225m) → k = 5333 N/m Now we can find the potential energy: Uelastic = ½ kx2 Uelastic = ½(5333 N/m)(0.035m)2 Uelastic = 3.27 J 6) A 100kg crate is pulled along a rough level floor at constant speed by a rope inclined at an angle of 30°. If the coefficient of kinetic friction is 0.4, find the tension in the rope and the work done by the rope’s tension when the crate moves 10m. This problem is mainly a review problem for chapter 5, with a bit of chapter 7 thrown in at the end. We can start by drawing the free-body diagram for the crate, and setting up the 2 versions of Newton’s 2 Normal friction Tension mg ∑ Fx = 0 Tx − fkinetic = 0 Law. ∑ Fy = 0 Constant speed means Fnet=0 N + Ty − mg = 0 T ⋅ cos(30 o ) = µ k ⋅ N N = mg − T sin(30 o ) ( T ⋅ cos(30 o ) = µ k ⋅ mg − T sin(30 o ) ( nd ) ) Substitute for N in the x-equation. T cos(30 o ) + µ k ⋅ sin(30 o ) = µ k ⋅ mg T= µ k ⋅ mg cos(30 o ) + µ k ⋅ sin(30 o ) = 368N Work = F ⋅ d ⋅ cos θ ⇒ W = (368N) ⋅ (10m) ⋅ cos(30 o ) = 3185J www.clas.ucsb.edu/staff/vince