Question #41008

Two forces are applied to a 2.0-kilogram block on a frictionless horizontal surface, as
shown in the diagram below. The acceleration of the block is

http://www.webassign.net/regents/rphys_jun00-12.gif

(a) 1.5 m/s2 to the right
(b) 2.5 m/s2 to the left
(c) 2.5 m/s2 to the right
(d) 4.0 m/s2 to the left
(e) non of the above

Expert's answer

Two forces are applied to a 2.0-kilogram block on a frictionless horizontal surface, as

shown in the diagram below. The acceleration of the block is http://www.webassign.net/regents/rphys_jun00-12.gif



(a) 1.5m/s21.5\mathrm{m / s2} to the right

(b) 2.5 m/s22.5 \mathrm{~m} / \mathrm{s} 2 to the left

(c) 2.5 m/s22.5 \mathrm{~m} / \mathrm{s} 2 to the right

(d) 4.0 m/s24.0 \mathrm{~m} / \mathrm{s} 2 to the left

(e) non of the above

Solution

Total force:


Ftotal=ΣF\overrightarrow {F _ {t o t a l}} = \Sigma \vec {F}


Take left as positive direction:


Ftotal=F1F2=83=5NF _ {t o t a l} = F _ {1} - F _ {2} = 8 - 3 = 5 N


According to the Second Newton's law:


F=maF = m * aa=Fm=52=2.5m/s2a = \frac {F}{m} = \frac {5}{2} = 2. 5 m / s ^ {2}


Answer: (b) 2.5m/s22.5 \, \text{m/s}^2 to the left

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