Answer :
The frictional force on the block from the floor and the block's acceleration are 10.45 N and 0.73 m/s².
Explanation :
Given that,
Mass of block = 3.50
Angle = 30°
Force = 15.0 N
Coefficient of kinetic friction = 0.250
We need to calculate the frictional force
Using formula of frictional force
[tex]F_{k}=\mu N[/tex]
[tex]F_{k}=\mu (F\sin\theta+mg)[/tex]
[tex]F_{k}=0.250\times(15\times\sin30^{\circ}+3.50\times9.8)[/tex]
[tex]F_{k}=0.250\times41.8[/tex]
[tex]F_{k}=10.45\ N[/tex]
(II). We need to calculate the block's acceleration
Using newton's second law of motion
[tex]F=ma[/tex]
[tex]a=\dfrac{F}{m}[/tex]
[tex]a=\dfrac{F\cos\theta-F_{k}}{m}[/tex]
[tex]a=\dfrac{15.0\cos30^{\circ}-10.45}{3.50}[/tex]
[tex]a=0.73\ m/s^2[/tex]
Hence, The frictional force on the block from the floor and the block's acceleration are 10.45 N and 0.73 m/s².
(a) The frictional force on the block is 10.45 N
(b) The acceleration of the block is 0.725 m/s²
The given parameters;
mass of the block, m = 3.5 kg
applied force, F = 15 N
angle of inclination θ = 30.0°
coefficient of kinetic friction, μ = 0.25
The resultant of vertical component of the for force on the block is calculated from Newton's second law of motion as shown below;
[tex]\Sigma F= 0\\\\F_n - 15sin(\theta) - mg = 0\\\\F_n = mg + 15sin(\theta)\\\\F_n = (3.5\times 9.8) + 15sin(30)\\\\F_n = 41.8[/tex]
(a) The frictional force on the block is calculated as;
[tex]F_k = u_kF_n\\\\F_k = 0.25 \times 41.8 \\\\F_k = 10.45 \ N[/tex]
(b) The acceleration of the block is calculated as;
[tex]Fcos(\theta ) - F_k = ma\\\\15 cos(30) - 10.45 = 3.5a\\\\12.99 - 10.45 = 3.5 a\\\\2.54 =3.5a\\\\a = \frac{2.54}{3.5} \\\\a = 0.725 \ m/s^2[/tex]
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