A half-full recycling bin has mass 3.0 kg and is pushed up a 40.0^\circ40.0 ∘ incline with constant speed under the action o
1 answer:
Answer:
11.82 N upward and parallel to the incline.
Explanation:
Let g = 9.81 m/s2. Gravity that acts on the bin has a magnitude of
W = mg = 3*9.81 = 29.43 N
The gravity component that is parallel to the 40 degree incline is
If the bin is moving at constant speed, then the net force (parallel to the incline) is also 0. Gravity is balanced with the push force of 26 N and the friction force.
The magnitude of friction force is 26 - 18.9 = 7.08 N
The the bin is moving down the incline, then gravity is 18.9N , friction is 7.08 N, one must create an upward force of 18.9 - 7.08 = 11.82 N parallel with the incline to balance this.
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Answer:
Friction force on the bullet is 58.7 N opposite to its velocity
Explanation:
As we know that initial speed of the bullet is 55 m/s
after travelling into the sand bag by distance d = 1.34 m it comes to rest
so final speed
now we can use kinematics top find the acceleration of the bullet
so we have
now by Newton's II law we know that
so we have
To solve this problem we will apply the linear motion kinematic equations. From the definition of the final velocity, as the sum between the initial velocity and the product between the acceleration (gravity) by time, we will find the final velocity. From the second law of kinematics, we will find the vertical position traveled.
Here,
v = Final velocity
= Initial velocity
g = Acceleration due to gravity
t = Time
At t = 4s, v = -30m/s (Downward)
Therefore the initial velocity will be
Now the position can be calculated as,
When it has the ground, y=0 and the time is t=4s,
Therefore the cliff was initially to 41.6m from the ground