Answer:
sure if its easy im jp ill help u
Explanation:
Answer:
a = 5.65 m/s² (↑)
Explanation:
T = 34.012 N (↑)
m = 2.2 Kg
g = 9.81 m/s²
We apply Newton's 2nd Law as follows:
∑ Fy = m*a
⇒ T - W = m*a
⇒ T - m*g = m*a
⇒ a = (T - m*g) / m
Finally
⇒ a = (34.012 N - 2.2 Kg*9.81 m/s²) / 2.2 Kg
⇒ a = 5.65 m/s² (↑)
Using SUVAT (S is displacement, U is initial Velocity, V is final velocity, A is acceleration, t is time);
S = 80m
U = 0ms^-1
V = ?
A = 9.8ms^-2
T = XXX
Use S U and A to find V;
V^2=U^2+2AS
V^2=0+2*80*9.8
V^2=1568
V=39.6ms^-1
Answer:
a) 4.98m/s²
b) 481.66N
Explanation:
a) Using the Newtons second law of motion
![\sum F_x = ma_x\\F_m - F_f = ma_x\\Wsin \theta - F_f = ma_x\\mgsin \theta - F_f = ma_x\\](https://tex.z-dn.net/?f=%5Csum%20F_x%20%3D%20ma_x%5C%5CF_m%20-%20F_f%20%3D%20ma_x%5C%5CWsin%20%5Ctheta%20-%20F_f%20%3D%20ma_x%5C%5Cmgsin%20%5Ctheta%20-%20F_f%20%3D%20ma_x%5C%5C)
m is the mass of the object
g is the acceleration due to gravity
Fm is the moving force acting along the plane
Ff is the frictional force opposing the moving froce
a is the acceleration of the skier
Given
m = 60kg
g = 9.8m/s²
= 35°
Ff = 38.5N
Required
acceleration of the skier a
Substituting into the formula;
![60(9.8)sin 35^0 - 38.5 = 60a\\588sin35^0 - 38.5 = 60a\\337.26 - 38.5 = 60a\\298.76 = 60a\\a = 298.76/60\\a = 4.98m/s^2\\](https://tex.z-dn.net/?f=60%289.8%29sin%2035%5E0%20-%2038.5%20%3D%2060a%5C%5C588sin35%5E0%20-%2038.5%20%3D%2060a%5C%5C337.26%20-%2038.5%20%3D%2060a%5C%5C298.76%20%3D%2060a%5C%5Ca%20%3D%20298.76%2F60%5C%5Ca%20%3D%204.98m%2Fs%5E2%5C%5C)
Hence the acceleration of the skier is 4.98m/s²
b) The normal force on the skier is expressed as;
N = Wcosθ
N = mgcosθ
N = 60(9.8)cos 35°
N = 588cos 35°
N = 481.66N
Hence the normal force on the skier is 481.66N
The formula for speed is speed = distance ÷ time. To work out what the units are for speed, you need to know the units for distance and time. In this example, distance is in metres (m) and time is in seconds (s), so the units will be in metres per second (m/s).