You pull a short refrigerator with a constant force across a greased (frictionless) floor, either with horizontal (case 1) or with tilted upward at an angle 38° (case 2). (a) What is the ratio of the refrigerator's speed in case 2 to its speed in case 1 if you pull for a certain time t? (b) What is this ratio if you pull for a certain distance d?

a. cos38

b. radical (cos38)

glad to help one of my classmates in AUB ;)

To solve this problem, we need to use basic principles of force and motion, particularly Newton's second law, along with the concept of work and energy.

(a) Ratio of speed in case 2 to case 1 when pulling for a certain time t:
In order to determine the ratio of speeds between case 2 and case 1, we first need to understand the forces at play in each case.

In case 1, when the refrigerator is pulled horizontally, the only force acting on it is the applied force in the pulling direction. Therefore, by Newton's second law (F = ma), we can write:

F1 = m * a1

Here, F1 is the applied force, m is the mass of the refrigerator, and a1 is the resulting acceleration.

In case 2, when the refrigerator is pulled upward at an angle of 38°, we have two forces to consider. The first force is the applied force in the upward direction, and the second force is the weight of the refrigerator, acting vertically downwards. We can resolve these forces into two components: one parallel to the floor and one perpendicular to the floor.

The force parallel to the floor causes the acceleration:

F2_parallel = m * a2_parallel

The force perpendicular to the floor cancels out the weight and prevents the refrigerator from sinking into the floor:

F2_perpendicular = m * g * cos(38°) = N

where g is the acceleration due to gravity.

The ratio of the speed in case 2 (v2) to the speed in case 1 (v1) is given by:

v2 / v1 = (a2_parallel / a1)

Now, let's calculate the acceleration in each case.

In case 1, the acceleration can be determined using the applied force and the mass of the refrigerator:

a1 = F1 / m

In case 2, the total force parallel to the floor can be calculated using the applied force and the component of the weight parallel to the floor:

F2_parallel = F_applied - F2_perpendicular

The acceleration in case 2 can then be found using:

a2_parallel = F2_parallel / m

Finally, we can determine the ratio of speeds (v2/v1) by substituting the respective expressions for acceleration into the equation above.

(b) Ratio of speed in case 2 to case 1 when pulling for a certain distance d:
To find the ratio of speeds when pulling for a certain distance d, we need to consider the work-energy principle.

The work done in case 1 can be calculated as the product of the applied force and the distance traveled:

Work1 = F1 * d

In case 2, since the refrigerator is being pulled up an inclined surface at an angle, we need to calculate the work done against gravity and the work done along the inclined surface.

The work done against gravity is equal to the change in potential energy:

Work2_gravity = m * g * h

where h is the vertical height that the refrigerator is lifted.

The work done along the inclined surface can be calculated as the product of the component of the applied force parallel to the surface and the distance traveled along the surface:

Work2_parallel = F2_parallel * d

The total work done in case 2 is then:

Work2 = Work2_gravity + Work2_parallel

The ratio of the speeds in case 2 to case 1 can be determined from the ratio of the work done in case 2 to case 1:

v2 / v1 = sqrt(Work2 / Work1)

By plugging in the expressions for the respective works, we can find the desired ratio.

Remember to convert all the units to consistent systems (e.g., SI units) before performing the calculations.