Suppose a rock falls from rest from a height of 100 meters and the only force acting on it is gravity. Find an equation for the velocity v(t) as a function of time, measured in meters per second. Hint What is the initial velocity of the rock?

Pulleys C and D in Figure are fastened together. Weights A and B are supported by ropes wound around the pulleys as shown. The radius for pulley C is 195 mm and the radius for pulley D is 129 mm. If weight A is accelerating downward at 1.8 m/s2, determine the linear acceleration of weight B. Provide your answer in m/s2 and round to the nearest two decimal places.

An electron moves with a constant horizontal velocity of 3.0 x 100 m/s and no initial vertical velocity as it enters a deflector inside a TV tube. The electron strikes the screen after traveling 17.0 cm horizontally and 40.0 cm vertically upward with no horizontal acceleration. What is the constant vertical acceleration provided by the deflector? (The effects of gravity can be ignored.) 1.4 x 10 m/s? 14 2.5 x 10 m/s2 14 1.2 x 10 m/s? 8.3 x 10 m/s?

The weight of an individual walking is 824N. The hill they are walking on has a slope of 10 degrees. Assuming they are in a static equilibrium what is their normal force?

An undamped spring-mass system contains a mass that weighs 8lb and a spring with spring constant 3lb/in. It is suddenly set in motion at t=0 by an external force of 7cos⁡(4t)lb.   Determine the position u, in feet, of the mass at any time t. Use 32ft/s2 as the acceleration due to gravity. Pay close attention to the units. u(t)  = ? ft

65. If a bucket of water spins about a vertical axis with constant angular velocity o (in radians per second), the water climbs up the side of the bucket until it reaches an equilibrium position (Figure 11). Two forces act on a particle located at a distance x from the vertical axis: the gravitational force -mg acting downward and the force of the bucket on the particle (transmitted indirectly through the liquid) in the direction perpendicular to the surface of the water. These two forces must combine to supply a centripetal force mo?x, and this occurs if the diagonal of the rectangle in Figure 11 is normal to the water's surface (i.e., perpendicular to the tangent line). Prove that if y = f(x) is the equation of the curve obtained by taking a vertical cross section through the axis, then -1/y' = -8/(@²x). Show that y = f(x) is a parabola. y =fr) mw?x mg FIGURE 11

Question b)

A hollow steel ball weighing 4 pounds is suspended from a spring. This stretches the spriny feet. The ball is started in motion from the equilibrium position with a downward velocity of 6 feet per second. The air resistance (in pounds) of the moving ball numerically equals 4 times its velocity (in feet per second). Suppose that aftert seconds the ball is y feet below its rest position. Find y in terms of t. (Note that the positive direction is down.) Take as the gravitational acceleration 32 feet per second per second.

A seasoned parachutist went for a skydiving trip where he performed freefall before deploying the parachute. According to Newton's Second Law of Motion, there are two forcës acting on the body of the parachutist, the forces of gravity (F,) and drag force due to air resistance (Fa) as shown in Figure 1. Fa = -cv ITM EUTM FUTM * UTM TM Fg= -mg x(t) UTM UT UTM /IM LTM UTM UTM TUIM UTM F UT GROUND Figure 1: Force acting on body of free-fall where x(t) is the position of the parachutist from the ground at given time, t is the time of fall calculated from the start of jump, m is the parachutist's mass, g is the gravitational acceleration, v is the velocity of the fall and c is the drag coefficient. The equation for the velocity and the position is given by the equations below: EUTM PUT v(t) = mg -et/m – 1) (Eq. 1.1) x(t) = x(0) – Where x(0) = 3200 m, m = 79.8 kg, g = 9.81m/s² and c = 6.6 kg/s. It was established that the critical position to deploy the parachutes is at 762 m from the ground…