1. A force F = (3.99xi + 2.86yj) N acts on an object as it moves in the x direction from the origin to x = 4.45 m. Calculate the work done on the object by the force.

Attemp:
since we only consider x direction i multiplied 3.99 by 4.45 but it didn't work out.

2. A 10.6 kg weather rocket generates a thrust of 226.0 N. The rocket, pointing upward, is clamped to the top of a vertical spring. The bottom of the spring, whose spring constant is 402.0 N/m, is anchored to the ground. Initially, before the engne is ignited, the rocket sits at rest on top of the spring.

A: After the engine is ignited, what is the rocket's speed when the spring has stretched 19.4 cm past its natural length?

B: What would be the rocket's speed after travelling the distance if it weren't tied down to the spring?

Attemp:
i used 1/2kx^2+Fthrust(d) = 1/skx^2 + mg(d) + 1/2mv^2 to solve (A) but i'm getting the wrong answer and i also used 3 other equations to solve this but they all wrong.
i have no idea how to start part (B)

Thank you

Thank You bobpursley.

I think i got it now.

1. To calculate the work done by a force, you need to find the dot product of the force vector and the displacement vector. In this case, the force vector is F = (3.99xi + 2.86yj) N and the displacement vector is Δx = 4.45 m in the x direction.

To find the dot product, you need to multiply the corresponding components of the two vectors and sum them up. In this case, the dot product is given by:

Work = F · Δx = (3.99)(4.45) + (2.86)(0) = 17.755 + 0 = 17.755 J

So, the work done on the object by the force is 17.755 Joules.

2. (A) To find the rocket's speed when the spring has stretched, you can use the principle of conservation of mechanical energy. The initial potential energy stored in the spring is converted into the final kinetic energy of the rocket.

The equation you used, 1/2kx^2 + F_thrust(d) = 1/2mv^2 + mg(d), is correct. Here's how you can solve it:

1/2kx^2 + F_thrust(d) = 1/2mv^2 + mg(d)

First, convert the displacement x = 19.4 cm to meters: x = 0.194 m.

Next, rearrange the equation to solve for the final velocity v:

1/2mv^2 = 1/2kx^2 + F_thrust(d) - mg(d)
v^2 = (kx^2 + 2F_thrust(d) - 2mg(d))/m
v = √[(kx^2 + 2F_thrust(d) - 2mg(d))/m]

Plug in the given values:
m = 10.6 kg
k = 402.0 N/m
x = 0.194 m
F_thrust(d) = 226.0 N
g = 9.8 m/s^2

Calculate v using the equation above, and you will find the speed of the rocket when the spring is stretched.

(B) To find the rocket's speed after traveling the distance if it weren't tied down to the spring, you can use the equation for projectile motion.

The energy equations used in part (A) are specific to the situation where the rocket is tied to the spring. However, if the rocket is not tied to the spring, you can treat it as a projectile and use the kinematic equations.

Use the equation: v^2 = u^2 + 2as

u is the initial velocity, which is zero since the rocket starts from rest on the spring.
a is the acceleration, which is due to gravity and is equal to g = 9.8 m/s^2.
s is the distance traveled, which you can calculate by considering the vertical displacement since the rocket is moving vertically.

Plug in the values, and you will find the speed of the rocket after traveling the distance.