Answer:
1.736m/s²
Explanation:
According to Newton's second law;
where;
Fm is the moving force = 70.0N
Ff is the frictional force acting on the body
is the coefficient of friction
m is the mass of the object
g is the acceleration due to gravity
a is the acceleration/deceleration
The equation becomes;
Substitute the given parameters
Hence the deceleration rate of the wagon as it is caught is 1.736m/s²
<h2>
Answer:</h2>
0.126m
<h2>
Explanation:</h2>
According to Hooke's law, the force (F) acting on a spring to cause an extension or compression (e) is given by;
F = k x e -------------------(i)
Where;
k = the spring's constant.
From the question, the force acting on the spring is the weight(W) of the mass. i.e
F = W -----------------------(ii)
<em>But;</em>
W = m x g;
where;
m = mass of the object
g = acceleration due to gravity [usually taken as 10m/s²]
<em>From equation (ii), it implies that;</em>
F = W = m x g
<em>Now substitute F = m x g into equation(i) as follows;</em>
F = k x e
m x g = k x e ------------------(iii)
<em>From the question;</em>
m = m1 = 3.5kg
k = 278N/m
<em>Substitute these values into equation (iii) as follows;</em>
3.5 x 10 = 278 x e
35 = 278e
<em>Now solve for e;</em>
e = 35/278
e = 0.126m
Therefore, the distance the spring is stretched from its unstretched length (which is the same as the extension of the spring) is 0.126m
120n
since the speed is doubled, her force is doubled
The acceleration would be 6m/sThis is because of the formula, "f/m=a" to find the acceleration; We would need to subtract the force of the friction which equals 1380, then divide that by the mass (which was 230) to get the answer 6
