What forces affect a toy car as it moves down a ramp?
2 answers:
You might be interested in
Answer:
a1 = 3.56 m/s²
Explanation:
We are given;
Mass of book on horizontal surface; m1 = 3 kg
Mass of hanging book; m2 = 4 kg
Diameter of pulley; D = 0.15 m
Radius of pulley; r = D/2 = 0.15/2 = 0.075 m
Change in displacement; Δx = Δy = 1 m
Time; t = 0.75
I've drawn a free body diagram to depict this question.
Since we want to find the tension of the cord on 3.00 kg book, it means we are looking for T1 as depicted in the FBD attached. T1 is calculated from taking moments about the x-axis to give;
ΣF_x = T1 = m1 × a1
a1 is acceleration and can be calculated from Newton's 2nd equation of motion.
s = ut + ½at²
our s is now Δx and a1 is a.
Thus;
Δx = ut + ½a1(t²)
u is initial velocity and equal to zero because the 3 kg book was at rest initially.
Thus, plugging in the relevant values;
1 = 0 + ½a1(0.75²)
Multiply through by 2;
2 = 0.75²a1
a1 = 2/0.75²
a1 = 3.56 m/s²
Answer:
+7.0 m/s
Explanation:
Let's take rightward as positive direction.
So in this problem we have:
a = -2.5 m/s^2 acceleration due to the wind (negative because it is leftward)
t = 4 s time interval
v = -3.0 m/s is the final velocity (negative because it is leftward)
We can use the following equation:
v = u + at
Where u is the initial velocity
We want to find u, so if we rearrange the equation we find:
and the positive sign means the initial direction was rightward.