Ablock rests on a horizontal frictionless surface. a string is attached to the block, and is pulled with a force of 46.0 n at an angle θ above the horizontal. after the block is pulled through a distance of 2.00 m, its speed is 3.10 m/s, and 42.0 j of work has been done on it. what is the angle θ?

The angle is 62.83 degrees

Explanation:

Hello!

Briefly, the concept of work is defined as the energy needed to move a certain distance body, it is calculated as the product of the force in the direction of movement by the distance traveled.

Taking into account the above, this exercise is resolved with the following steps.

1. draw a problem diagram involving the forces and angle (see attached image)

2. As you can see only the horizontal component (fcos) is doing work, then the equation for the work of this body is given by the horizontal force by the distance traveled as follows

W=FcosΘd

Where

W=work=42J

F=force=46N

d=distance traveled=2m

now we use algebra to find the angle, and use the values of force, distance, and work

The angle is 62.83 degrees

the second one and the forth one are in the purple because they are opinions and the rest are in the yellow

explanation:

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s