HELP. exam tomorrow. practice midterm questions

[taproot729o6]

Two problems.
Problem 1. (30 Points)

When leaving for work, a woman leaves a cup of coffee on top of her car. The cup stays
fixed on the car, and the car moves according to the function:

x(t) = (2.0 m/s) t + (1.0 m/s2) t2

At time t = 2.0 s, a gust of wind comes from behind the car and blows the cup of coffee
off the car with a velocity of v = 5 m/s at an angle θ = 30° from horizontal. The coffee
cup strikes the ground after some time.

(a) What are the magnitudes of the velocities of (i) the car and (ii) the coffee when
the coffee flies off the car? (20 Points)
(b) Where does the coffee land with respect to the car? (Take the initial position of
the coffee to be the very front of the car.) (10 Points)



Problem 2. (30 Points)

A block with m=1.00 Kg sits d= 5.00 m from the edge of a platform. A constant
force of magnitude F=50.0 N is applied to the block at the angle of θ = 10° below
the horizontal. The applied force stops when the block reaches the edge of the
platform. The block lands a distance of l = 15.0 m from the platform. What is the
height of the platform, h? Assume all the surfaces are frictionless.

 

[alphysicist]

Hi taproot729o6,

From the thread title, you may not be interested in these problems anymore. If you are, please show what you have tried so far, so that we can see how to help.

 

[thomate1]

For Problem 1:
(a) (i) The magnitude of the car's velocity is given by the function x(t), so at t=2.0 s, the car's velocity is 6 m/s. (ii) The magnitude of the coffee's velocity can be found by using the given velocity and angle to calculate the horizontal and vertical components. The horizontal component is 5*cos(30°) = 4.33 m/s and the vertical component is 5*sin(30°) = 2.5 m/s. Therefore, the magnitude of the coffee's velocity is √(4.33^2 + 2.5^2) = 5.09 m/s.

(b) The coffee will land a horizontal distance of 6 m from the front of the car, since it travels at a constant horizontal velocity of 4.33 m/s for 2 seconds before being blown off the car.

For Problem 2:
To solve this problem, we can use the equations of motion for constant acceleration. The first equation, x = x0 + v0t + 1/2at^2, can be used to find the time it takes for the block to reach the edge of the platform. Since x0 = 5 m, v0 = 0 m/s, and a = 9.8 m/s^2 (since the block is falling under the force of gravity), we can rearrange the equation to solve for t. Plugging in the values, we get t = 1.02 s.

Next, we can use the second equation, v = v0 + at, to find the initial velocity of the block. Since v = 0 m/s (since the block stops at the edge of the platform), v0 = 9.8*1.02 = 10 m/s.

Finally, we can use the third equation, v^2 = v0^2 + 2aΔx, to solve for the height of the platform. Plugging in the values, we get 0^2 = 10^2 + 2*(-9.8)*h. Solving for h, we get h = 5.1 m.

Therefore, the height of the platform is approximately 5.1 m.

 

[baba89]

I understand the stress and pressure of an upcoming exam. It is important to remember to stay calm and focused in order to perform well. In regards to the practice midterm questions, it is crucial to use them as a tool to assess your understanding and identify any areas that need extra attention.

For problem 1, it is clear that the car is moving with a constant acceleration of 1.0 m/s^2. At t=2.0 s, the car will have a velocity of 4.0 m/s and the coffee will have a velocity of 5 m/s at an angle of 30 degrees from horizontal. The coffee will land at a distance of 10.77 m from the car, calculated using the kinematic equations.

For problem 2, we can use the kinematic equations to find the velocity of the block when it reaches the edge of the platform, which is 9.49 m/s. Using this velocity and the given distance, we can find the time it takes for the block to reach the edge, which is 1.58 seconds. Using this time and the given angle, we can find the height of the platform to be 8.35 m.

Remember to approach these problems systematically and use the given information to your advantage. Best of luck on your exam tomorrow.