A force F = (6.60î − 3.65t ĵ) N, with t in units of s, is applied to a 2.00 kg object initially at rest.

Question

a) At what time (in s) will the object be moving with a speed of 15.0 m/s? (Round your answer to at least two decimal places.)

b) How far (in m) is the object from its initial position when its speed is 15.0 m/s?

c) What is the total displacement (in m) through which the object traveled at this time? (Express your answer in vector form.)

Answer 1

Let's break up the vector into its horizontal and vertical components. So we have a force of 6.60 N in the x-direction and a force of -3.65 N in the y-direction (so directed downward).

For part a), we are dealing with speed, which is a scalar, so we can calculate the magnitude of the force as the square root of the sum of the squares of each of the forces. so the square root of (6.60 N )² + (-3.65 N )² is

7.542 N.

Since F = ma, the acceleration is F/m = 7.542 N / 2.00 kg = 3.7710 m/sec².

To get the time at which the object reaches a speed of 15 m/sec, we use the equation for an object accelerating from rest, v = at, so t = v/a = (15.0 m/sec) / (3.7710 m/sec²) = 3.9777 sec

For part b), we are looking for the distance traveled in this time.

There are two approaches here:

We can use the traditional equation for a body accelerating from rest, d = 1/2 at²,

so (1/2)(3.7710 m/sec²)(3.9777sec)² = 29.8325 m

A second approach would be to realize that the average velocity over the interval is 15.0 m/sec divided by 2, or 7.5 m/sec, since the object is uniformly accelerating from rest.

So multiplying 7.5 m/sec by the 3.9777 sec gives the same answer of 29.8325 m.

For c), we want the displacement in vector form. Since we already have the total displacement, this would be the magnitude of the vector, which is 29.8325 m. To get the components of each force, we need the angle of the force.

Realizing that the angle of the force is determined by its x and y components, we can draw a vector sum triange, where the x value is 6.60 N and the y value is -3.65 N, so the tangent of the angle separating the vectors is (6.60 N)/ (-3.65N) = -0.5530, so the angle in degrees is

arctan (-0.5530) = -28.9439 degrees (so the vector goes to the right and downward).

The x-component of the displacement would then be 29.8325 m times the cosine of the angle above, or 29.8325 m (cos (-28.9439)) = 26.1062 m, and the y component given by the sine of the angle times the magnitude, which turns out to be 29.8325 m (sin (-28.9439)) = -14.4375 m

So, the final vector of displacement is 26.062 i - 14.4375 j.

A good check here would be to verify that the ratio of the displacement components matches the ratio of the force components. See if they indeed match up.

I hope that helps Ramitha. And please verify the math here!

Stephen

Answer 2

You have a constant force at i direction and a force proportional to t at j direction.

a)

on i direction: acceleration a_i = 6.6/2=3.3 (m/s²), v_i = a_i*t=3.3t (m/s)

on -j direction: acceleration a_j= 3.65t/2=1.825t (m/s²), v_j=INT(a_j*dt)=0.91t²(m/s)

(3.3t)² + (0.91t²)²=15²

Solve using calculator. t²=11.16, t=3.3s

b)

same approach.

on i direction, constant acceleration, S_i = a_it²/2 = 1.65*11.16=18.42 (m)

on -j direction, S_j = INT(v_jdt) =0.3*t³=10.9(m)

total distance: sqrt(S_i²+S_j²) =21.4(m)

c)

18.42i -10.9j