Question

Space scientists have a large test chamber from which all the air can be evacuated and in which they can create a horizontal uniform electric field. The electric field exerts a constant hor-izontal force on a charged object. A 15 g charged projectile is launched with a speed of 6.0 m/s at an angle 35° above the hori-zontal. It lands 2.9 m in front of the launcher. What is the magni-tude of the electric force on the projectile?

Answer 1

Answer:

Magnitude of electric force = 0.03345 N

Explanation:

We are given;

Mass; m = 15g = 0.015kg

Angle above horizontal; θ = 35°

Speed; v = 6 m/s

Horizontal displacement; d = 2.9m

Now formula for time of flight is given as;

time of flight; t = (2Vsinθ)/g

Thus, plugging in values, we have

t = (2 x 6.0 x sin35)/9.8

t = (12 x 0.5736)/9.8

t = 0.7024 s

Now, let's find the acceleration

The formula for horizontal displacement is given by;

d = (Vcosθ)t + (1/2)at²

Plugging in the relevant values ;

2.9 = [6(cos35) x 0.7024] + (1/2)a(0.7024)²

2.9 = (4.2144 x 0.8192) + (0.2467)a

2.9 = 3.45 + (0.2467)a

(0.2467)a = 2.9 - 3.45

a = -0.55/0.2467

a = -2.23 m/s²

Since we are looking for the magnitude of the electric force, we will take the absolute value of a. Thus, a = 2.23 m/s²

We know that F = ma

Thus,Force = 0.015kg x 2.23m/s² =

= 0.03345 N

Answer 2

Answer:

the magnitude of the electric force on the projectile is 0.0335N

Explanation:

time of flight t = 2·V·sinθ/g

= (2 * 6.0m/s * sin35º) / 9.8m/s²

= 0.702 s

The body travels for this much time and cover horizontal displacement x from the point of lunch

So, use kinematic equation for horizontal motion

horizontal displacement

x = Vcosθ*t + ½at²

2.9 m = 6.0m/s * cos35º * 0.702s + ½a * (0.702s)²

a = -2.23 m/s²

This is the horizontal acceleration of the object.

Since the object is subject to only electric force in horizontal direction, this acceleration is due to electric force only

Therefore,the magnitude of the electric force on the projectile will be

F = m*|a|

= 0.015kg * 2.23m/s²

= 0.0335 N

Thus, the magnitude of the electric force on the projectile is 0.0335N