Answer:
The crate was being lifted by a height of 1.48 meters.
Explanation:
In an attempt o move a crate;
Force applied = 2470 N
Work done by the force = 3650 J
We know that the work done is defined as the force used to move an object to a distance.
Given the Force used and the work done by that Force, we need to find out the distance the crate was lifted to.
Work done is defined as:
Work = Force*distance covered in the direction of the force
3650 = 2470*distance
distance = 3650/2470
distance = 1.48 meters
The parcel will undergo projectile motion, which means that it will have motion in both the horizontal and vertical direction.
First, we determine how long the parcel will fall using:
s = ut + 1/2 at²
where s will be the height, u is the initial vertical velocity of the parcel (0), t is the time of fall and a is the acceleration due to gravity.
5.5 = (0)(t) + 1/2 (9.81)(t)²
t = 1.06 seconds
Now, we may use this time to determine the horizontal distance covered by the parcel by using:
distance = velocity * time
The horizontal velocity of the parcel will be equal to the horizontal velocity of the cruise liner.
Distance = 10 * 1.06
Distance = 10.6 meters
The boat should be 10.6 meters away horizontally from the point of release.
Given
m1(mass of red bumper): 225 Kg
m2 (mass of blue bumper): 180 Kg
m3(mass of green bumper):150 Kg
v1 (velocity of red bumper): 3.0 m/s
v2 (final velocity of the combined bumpers): ?
The law of conservation of momentum states that when two bodies collide with each other, the momentum of the two bodies before the collision is equal to the momentum after the collision. This can be mathemetaically represented as below:
Pa= Pb
Where Pa is the momentum before collision and Pb is the momentum after collision.
Now applying this law for the above problem we get
Momentum before collision= momentum after collision.
Momentum before collision = (m1+m2) x v1 =(225+180)x 3 = 1215 Kgm/s
Momentum after collision = (m1+m2+m3) x v2 =(225+180+150)x v2
=555v2
Now we know that Momentum before collision= momentum after collision.
Hence we get
1215 = 555 v2
v2 = 2.188 m/s
Hence the velocity of the combined bumper cars is 2.188 m/s
