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Help with Question #2 and please show work. Measured the position versus time by

ID: 2118237 • Letter: H

Question



Help with Question #2 and please show work.

Measured the position versus time by using a falling mass to pull the cart along the track. The data obtained are shown below. If the relationship between the cart's position and time were known to be x(t) = x0[1 + kt(3/2)], construct a straight-line graph of the appropriate variables and use the graph to determine the values for x0 and k. Include units. (b) At what time is the cart located at 18.7 cm? (c) Where will be cart be at 4.6 seconds? Please show all work on the graph. Using the same force table that you used, a student was given two forces and asked to find a single force that would produce equilibrium with the two forces. The two forces were given by providing one of the components and the angle the force made with the positive x axis. The two forces were as follows: Force 1: Fx = -2.7N; theta = 220 degrees Force 2: Fy = 3.2 N; theta = 6 degrees For each force, calculate the magnitude of the force and the mass required to produce it. (b) Calculate the mass and its angle of application to produce equilibrium with these two forces. Please specify your angle as counterclockwise positive with respect to the positive x axis. A very light spring is 10.0 cm long with no mass attached to it. When 50.0 grams arc added to it. its length increases to 11.25 cm. Another 100.0-g mass is added to the and it is set into oscillation How much time is required for 100 complete oscillations?

Explanation / Answer

force 1 (x component) : -2.7N angle = 220 degrees so total force would be -2.7/cos(220) = 3.52N

force 2 (y component): 3.2N angle 60 degrees ; so total force would be 3.2/sin(60) = 3.69N


y component of force 1 = 3.52*sin(220) = -2.26N

x component of force 2 - 3.69*cos(60) = 1.85N


total x component of both forces = - 0.85N

total y component of both forces = 0.94N


so force required for quilibrium is opposite to total force : so x component = 0.85N ; y component = -0.94N

this force is in the 4th quadrant and angle made with x axis is 360-48 = 312 degrees counterclockwise from x axis

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