A 465 g block slides along a frictionless surface at a speed of 0.35 m/s. It runs into a horizontal massless spring with spring
1 answer:
Answer:
a) The duration, during which the block remain in contact with the spring is 0.29 s
b) The period of the simple harmonic oscillatory motion depends only on the mass and spring constant, therefore when the speed is doubled, the duration of contact remains the same as 0.29 s.
Explanation:
Mass of the block = 465 g
Surface speed = 0.35 m/s
Spring constant , k = 54 N/m
= 0.58 s
a) Since the period for the oscillatory motion is 0.58 s, then the time when the block and spring remain in contact is T/2 = 0.29 s
b) When the speed is doubled, we have
Therefore, since T is only dependent on the mass, m and the spring constant, K, then the time it takes when the speed is doubled remain as
T /2 = 0.29 s
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Answer:
The sled required 9.96 s to travel down the slope.
Explanation:
Please, see the figure for a description of the problem. In red are the x and y-components of the gravity force (Fg). Since the y-component of Fg (Fgy) is of equal magnitude as Fn but in the opposite direction, both forces get canceled.
Then, the forces that cause the acceleration of the sled are the force of the wind (Fw), the friction force (Ff) and the x-component of the gravity force (Fgx).
The sum of all these forces make the sled move. Finding the resulting force will allow us to find the acceleration of the sled and, with it, we can find the time the sled travel.
The magnitude of the friction force is calculated as follows:
Ff = μ · Fn
where :
μ = coefficient of kinetic friction
Fn = normal force
The normal force has the same magnitude as the y-component of the gravity force:
Fgy = Fg · cos 30º = m · g · cos 30º
Where
m = mass
g = acceleration due to gravity
Then:
Fgy = m · g · cos 30º = 87.7 kg · 9.8 m/s² · cos 30º
Fgy = 744 N
Then, the magnitude of Fn is also 744 N and the friction force will be:
Ff = μ · Fn = 0.151 · 744 N = 112 N
The x-component of Fg, Fgx, is calculated as follows:
Fgx = Fg · sin 30º = m·g · sin 30º = 87.7 kg · 9.8 m/s² · sin 30º = 430 N
The resulting force, Fr, will be the sum of all these forces:
Fw + Fgx - Ff = Fr
(Notice that forces are vectors and the direction of the friction force is opposite to the other forces, then, it has to be of opposite sign).
Fr = 161 N + 430 N - 112 N = 479 N
With this resulting force, we can calculate the acceleration of the sled:
F = m·a
where:
F = force
m = mass of the object
a = acceleration
Then:
F/m = a
a = 479N/87.7 kg = 5.46 m/s²
The equation for the position of an accelerated object moving in a straight line is as follows:
x = x0 + v0 · t + 1/2 · a · t²
where:
x = position at time t
x0 = initial position
v0 = initial velocity
t = time
a = acceleration
Since the sled starts from rest and the origin of the reference system is located where the sled starts sliding, x0 and v0 = 0.
x = 1/2· a ·t²
Let´s find the time at which the position of the sled is 271 m:
271 m = 1/2 · 5.46 m/s² · t²
2 · 271 m / 5.46 m/s² = t²
<u>t = 9.96 s </u>
The sled required almost 10 s to travel down the slope.
