A ship is launched toward the water on a slipway making an angle of 5 degree with horizontal. The coefficient of kinetic frictio
1 answer:
Newton's second law we can find the acceleration of the ship is
a = 0.074 m / s²
Given parameters
- The very coefficient of kinematic friction = 0.08
- The angle of the ramp tea = 5º
To find
- The acceleration of the ship
Newton's second law states that the force varies linearly with the mass and acceleration of the body
∑ F = m a
The bold indicate vectors, where F is the sum of the forces, m the masses and the acceleration
The force is a vector magnitude, therefore it must be solved for each axis of a coordinate system, in the exerted of inclined ramps the coordinate system used has the x-axis in the direction of the ramp and the y-axis perpendicular to it, see attached. In this cordinate system the weight of the body must be decomposed.
cos 5 = W_y / W
sin 5 = Wₓ / W
W_y = W cos 5
Wₓ = W sin 5
Let's solve Newton's second law for each axis
Y axis
N - W_y = 0
N = W_y = mg cos 5
X axis
Wₓ - fr = m a (1)
Friction force is the macroscopic result of contact interactions between two bodies and is described for solids by the expression
fr = μ N
fr = μ mg cos 5
we substitute in equation 1
mg sin 5 - μ m g cos 5 = m a
a = g (sin 5 - μ cos 5)
let's calculate
a = 9.8 (sin 5 - 0.08 cos 5)
a = 0.073 m / s²
In conclusion using Newton's second law we can find the acceleration of the ship is
a = 0.073 m / s²
Learn more about Newton's second law here:
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Answer:
a) 0.658 seconds
b) 0.96 inches
Explanation:
Time taken by the ball to reach the highest point is 0.14 seconds
The highest point reached by the snowball above its release point is 0.315 ft
Total height the snowball will fall is 4+0.315 = 4.315 ft
The snowball will reach the bank at 0.14+0.518 = 0.658 seconds after it has been thrown
The snowball goes 0.5-0.42 = 0.08 ft = 0.96 inches
Answer:
It traveled 4 centimeters.
Explanation:
In a speed versus time graph, the distance travelled is given by the area under the graph.
In this graph we have the following:
- The speed of the object is v = 1 cm/s between time t = 0 s and t = 4 s
- The speed of the object is v = 0 cm/s between time t = 4 s and t = 8 s
Since the speed in the second part is zero, the distance travelled in the second part is zero. So, the only distance travelled by the object is the distance travelled during the first part, which is equal to the area of the first rectangle: